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GRE Subject Test: Chemistry : General Chemistry

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Example Questions

Example Question #1 : Reaction Chemistry

$Na_2SO_4+BaCl_2 \rightarrow 2NaCl +BaSO_4$

When 15.5 grams of $Na_{2}SO_{4}$ is used in the given reaction, how many moles of $BaSO_{4}$ is produced?

Possible Answers:

4.122mol

0.1092mol

0.500mol

0.0278mol

Correct answer:

0.1092mol

Explanation:

We must first calculate the molecular weight of $Na_{2}SO_{4}$ to use as a conversion factor:

$MW_{Na_{2}SO_{4}}=(\frac{23g}{mole}\times 2)+(\frac{32g}{mole}\times 1)+(\frac{16g}{mole})=142\frac{g}{mole}$ $15.5g \times \frac{mole}{142g}=0.1092\ moles\ Na_{2}SO_{4}$

For every mole of $Na_{2}SO_{4}$ reacted, one mole of $BaSO_{4}$ is produced.

Therefore, 0.1092 moles of $BaSO_{4}$ is produced.

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Example Question #2 : Reaction Chemistry

$Na_2SO_4+BaCl_2 \rightarrow 2NaCl +BaSO_4$

When 20.0 grams of $Na_{2}SO_{4}$ is used in the given reaction, how many moles of NaCl is produced?

Possible Answers:

0.2183mol

3.22mol

1.223mol

0.1252mol

Correct answer:

0.2183mol

Explanation:

We must first calculate the molecular weight of $Na_{2}SO_{4}$ to use as a conversion factor:

$MW_{Na_{2}SO_{4}}=(\frac{23g}{mole}\times 2)+(\frac{32g}{mole}\times 1)+(\frac{16g}{mole})=142\frac{g}{mole}$ $15.5g \times \frac{mole}{142g}=0.1092\ moles\ Na_{2}SO_{4}$

For every 1 mole of $Na_{2}SO_{4}$ reacted, 2 moles of NaCl is produced.

Therefore,the number of moles of NaCl produced will be double the number of moles of $Na_{2}SO_{4}$ used up in the reaction:

$Moles\ of\ NaCl\ produced = 0.1092\ moles\ Na_{2}SO_{4}\times2=0.2183\ moles$

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Example Question #1 : Chemical Equations

$H_3PO_4+Ca(OH)_2\rightarrow H_2O+CaHPO_4$

What is the smallest whole-number coefficient that can be designated to the water molecule that would balance the chemical equation given?

Possible Answers:

Correct answer:

Explanation:

The number of atoms on the reactant side of the chemical equation are:

$Total\ O=6,\ Total\ P=1,\ Total\ H=5,\ Total\ Ca=1$

The number of atoms on the product side of the chemical equation are:

$Total\ O=5,\ Total\ P=1,\ Total\ H=3,\ Total\ Ca=1$

In a balanced chemical equation, the total number of atoms of each element must be equal on the reactant and product side of the chemical equation. There is one more oxygen and two more hydrogen atoms on the reactant side of the chemical equation. Giving the water molecule on the product side of the chemical equation would balance the chemical equation so that both sides contain six oxygens and five hydrogens: $H_3PO_4+Ca(OH)_2\rightarrow 2H_2O+CaHPO_4$

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Example Question #141 : General Chemistry

$C_3H_8+O_2\rightarrow CO_2+H_2O$

What is the smallest whole number coefficient that can be designated to the CO_2 molecule that would balance the chemical equation given?

Possible Answers:

Correct answer:

Explanation:

The process of balancing a chemical equation is called "balancing by inspection" which is done by trial and error. The easiest way to approach the problem given is by balancing the species containing an element with the most complicated formula occurring in one reactant and one product. In this case, we can either begin by balancing the carbon or hydrogen in the propane molecule. Let's start with the carbon by adding a coefficient of 3 to the carbon dioxide molecule on the product side of the equation which gives:

$C_{3}H_{8}\ +\ O_{2}\rightarrow\ 3CO_{2}\ +\ H_{2}O$

Let's now balance the hydrogens by adding a coefficient of 4 to the product side of the equation:

$C_{3}H_{8}\ +\ O_{2}\rightarrow\ 3CO_{2}\ +\ 4H_{2}O$

Lastly, we can balance the number of oxygens in the reaction:

$C_{3}H_{8}\ +\ 5O_{2}\rightarrow\ 3CO_{2}\ +\ 4H_{2}O$

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Example Question #10 : Chemical Equations

$NH_{3}(g)\ + HCl(l)\rightarrow\ NH_{4}Cl(s)$

If 0.150mol of $NH_{3}$ is used for the reaction given, how many grams of $NH_{4}Cl$ was formed?

Possible Answers:

0.500g

6.50g

0.125g

7.95g

26.5g

Correct answer:

7.95g

Explanation:

Based on the molecular equation, for every 1 mole of $NH_{3}$ reacted, 1 mole of $NH_{4}Cl$ is formed as a product. Using dimensional analysis, we can use this relationship to determine the moles of $NH_{3}$ reacted:

$0.150\ moles\ of\ NH_{3}\times\frac{1\ mole\ NH_{4}Cl}{1\ mole\ NH_{3}}=0.150\ moles\ of\ NH_{4}Cl$

We must calculate the molecular weight of $NH_{4}Cl$ :

$MW_{NH_{4}Cl}=(14\frac{g}{mole}\times 1)+(1\frac{g}{mole}\times 4)+(35\frac{g}{mole}\times1)=53\frac{g}{mole}$

To calculate the number of grams of $NH_{4}Cl$ , we need to convert the number of moles of $NH_{4}Cl$ to grams using the molecular weight as a conversion factor.

$0.150\ moles\ of\ NH_{4}Cl\ \times 53\frac{g}{mole}=7.95\ g$

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Example Question #11 : Reaction Chemistry

$CH_{4}\ + 4S\rightarrow\ CS_{2}\ +2H_{2}S$

If 0.250 mol of is used for the reaction given, how many grams of $H_{2}S$ was formed?

Possible Answers:

8.5g

4.25g

0.125g

2.13g

1.25g

Correct answer:

4.25g

Explanation:

Based on the molecular equation, for every 4 moles of reacted, 2 moles of $H_{2}S$ is formed as a product:

$0.250\ moles\ of\ S\times\frac{2\ mole\ H_{2}S}{4\ mole\ S}=0.125\ moles\ of\ H_{2}S$

We must calculate the molecular weight of $H_{2}S$ :

$MW_{H_{2}S}=(1\frac{g}{mole}\times 2)+(32\frac{g}{mole}\times 1)=34\frac{g}{mole}$

To calculate the number of grams of $H_{2}S$ , we need to convert the number of moles of $H_{2}S$ to grams using the molecular weight as a conversion factor.

$0.125\ moles\ of\ NH_{4}Cl\ \times 34\frac{g}{mole}=4.25\ g$

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Example Question #12 : Reaction Chemistry

$Fe_{2}O_{3}(aq)\ +2Al(s)\rightarrow\ 2Fe(s)+Al_{2}O_{3}(aq)$

If 0.340 mol of is formed during the reaction given, how many grams of $Fe_{2}O_{3}$ was used?

Possible Answers:

12.5g

27.2g

54.4g

0.122g

118g

Correct answer:

27.2g

Explanation:

Based on the molecular equation, for every 2 moles of formed, 1 mole of $Fe_{2}O_{3}$ is reacted:

$0.340\ moles\ of\ Fe\times\frac{1\ mole\ Fe_{2}O_{3}}{2\ mole\ Fe}=0.170\ moles\ of\ Fe_{2}O_{3}$

We must calculate the molecular weight of $Fe_{2}O_{3}$ :

$MW_{Fe_{2}O_{3}}=(56\frac{g}{mole}\times 2)+(16\frac{g}{mole}\times 3)=160\frac{g}{mole}$

To calculate the number of grams of $Fe_{2}O_{3}$ , we need to convert the number of moles of $Fe_{2}O_{3}$ to grams using the molecular weight as a conversion factor.

$0.170\ moles\ of\ NH_{4}Cl\ \times 160\frac{g}{mole}=\ 27.2\ g$

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Example Question #13 : Reaction Chemistry

$PCl_{3}(l)+3H_{2}O(l)\rightarrow\ H_{3}PO_{4}(l)+3HCl(l)$

If 0.160mol of HCl is formed during the reaction given, how many grams of $PCl_{3}$ was used?

Possible Answers:

0.125g

2.40g

0.500g

21.6g

7.21g

Correct answer:

7.21g

Explanation:

Based on the molecular equation, for every 3 moles of HCl formed, 1 mole of $PCl_{3}$ is reacted:

$0.160\ moles\ of\ HCl\times\frac{1\ mole\ PCl_{3}}{3\ mole\ HCl}=0.053\ moles\ of\ PCl_{3}$

We must calculate the molecular weight of $PCl_{3}$ :

$MW_{PCl_{3}}=(31\frac{g}{mole}\times 1)+(35\frac{g}{mole}\times 3)=136\frac{g}{mole}$

To calculate the number of grams of $PCl_{3}$ , we need to convert the number of moles of $PCl_{3}$ to grams using the molecular weight as a conversion factor.

$0.053\ moles\ of\ PCl_{3}\ \times 136\frac{g}{mole}=\ 7.21\ g$

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Example Question #1 : Oxidation Reduction Reactions

$MnO_{4}^{-}_{(aq)}+5Fe^{2+}_{(aq)}+8H^{+}_{(aq)}\rightarrow5Fe^{3+}_{(aq)}+Mn^{2+}_{(aq)}+4H_{2}O_{(l)}$

Based on the chemical equation given, which of the following compounds underwent reduction?

Possible Answers:

$Fe^{3+}$

$Mn^{7+}$

$O^{2-}$

$Fe^{2+}$

$H^{+}$

Correct answer:

$Mn^{7+}$

Explanation:

The type of chemical reaction given is called an oxidation-reduction reaction. In this type of reaction, there is a transfer of electrons from one element to another. The species gaining an electron is said to be reduced and the species losing an electron is oxidized. Manganese ion goes from a to a oxidation state. Therefore, manganese is reduced in the chemical reaction because it gains electrons from the compound being oxidized (iron).

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Example Question #1 : Oxidation Reduction Reactions

Methane combusts in the presence of oxygen according to the following reaction:

$CH_4 + 2O_2 \to CO_2 + 2H_2O$

Which of the following statements is true concerning the reaction?

Possible Answers:

Oxygen has a charge of $\small -2$ throughout the entire reaction

Oxygen is oxidized in the reaction

Carbon has an initial oxidation state of $\small +4$

Carbon is oxidized in the reaction

CH_4 will be the limiting reagent

Correct answer:

Carbon is oxidized in the reaction

Explanation:

By comparing the oxidation number of an atom as a reactant and its oxidation number as a product, we can determine if the atom has been oxidized or reduced. In increase in oxidation number indicates a loss of electrons, or oxidation. A decrease in oxidation number signals a gain of electrons, or reduction.

For electrochemistry, you should familiarize yourself with the traditional oxidation states of hydrogen $\small (+1)$ , halogens $\small (-1)$ , oxygen $\small (-2)$ , and elemental atoms $\small (0)$ .

Carbon is initially in the form of methane, meaning that it is attached to four hydrogen atoms. The molecule is neutral, and each hydrogen has an oxidation number of $\small +1$ . Carbon must have an initial oxidation state of $\small -4$ in order to balance the molecular charge.

In the a product, carbon is attached to two oxygens, each with a charge of $\small -2$ . Again, the molecule is neutral, so carbon must balance these charges. This means that carbon's final oxidation state is $\small +4$ .

Since carbon went from an oxidation state of $\small -4$ to $\small +4$ , we can conclude that carbon has been oxidized in the reaction.

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GRE Subject Test: Chemistry : General Chemistry

Study concepts, example questions & explanations for GRE Subject Test: Chemistry

All GRE Subject Test: Chemistry Resources

Example Questions

Example Question #1 : Reaction Chemistry

Example Question #2 : Reaction Chemistry

Example Question #1 : Chemical Equations

Example Question #141 : General Chemistry

Example Question #10 : Chemical Equations

Example Question #11 : Reaction Chemistry

Example Question #12 : Reaction Chemistry

Example Question #13 : Reaction Chemistry

Example Question #1 : Oxidation Reduction Reactions

Example Question #1 : Oxidation Reduction Reactions

All GRE Subject Test: Chemistry Resources

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