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MCAT Physical : Reaction Calculations and Limiting Reagent

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Example Question #21 : Reaction Calculations And Limiting Reagent

Consider the hydrolysis of acetyl chloride into acetic acid and hydrochloric acid:

$CH_3OCl_(_a_q_) + H_2O_(_l_) \rightleftharpoons CH_3COOH _(_a_q_) + HCl_(_a_q_)$

If 1g of acetyl chloride is added to water, how much hydrochloric acid is produced?

Possible Answers:

0.654g

0.231g

0.876g

0.464g

Correct answer:

0.464g

Explanation:

The limiting reagent in this reaction is the acetyl chloride (because water is never the limiting reagent; there is always an excess of water). To solve this question, we need to convert the mass of acetyl chloride to moles. The molecular weight of acetyl chloride is:

$12\frac{g}{mol} + (3)\left(1.008\frac{g}{mol}\right)+12\frac{g}{mol}+16\frac{g}{mol}+35.45\frac{g}{mol}$

$= 78.49\frac{g}{mol}$

The moles of acetyl chloride is:

$1g\left(\frac{1\: mol}{78.49g}\right) = 0.013\: mol$

The moles of hydrochloric acid produced is:

$(0.013\: mol) \frac{1\: mol\: HCl}{1\: mol} = 0.013 \: mol\:HCl$

The molecular weight of HCl is equal to $36.46\frac{g}{mol}$ . The mass of HCl produced is:

$0.013mol\left(\frac{36.46g}{1mol}\right) = 0.464g\: HCl$

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Example Question #22 : Reaction Calculations And Limiting Reagent

1g of A is added to 1g of B and the following reaction occurs.

$A_(_a_q_) + B_(_a_q_) \rightleftharpoons C_(_a_q_) + D_(_a_q_)$

The total amount of products produced is 0.016mol. What is the identity of A and B and what is the limiting reagent?

Possible Answers:

Molecule A: sodium chloride

Molecule B: potassium iodide

Limiting reagent: sodium chloride

Molecule A: sodium chloride

Molecule B: potassium iodide

Limiting reagent: potassium iodide

Molecule A: sodium chloride

Molecule B: potassium bromide

Limiting reagent: potassium bromide

Molecule A: sodium chloride

Molecule B: potassium bromide

Limiting reagent: sodium chloride

Correct answer:

Molecule A: sodium chloride

Molecule B: potassium bromide

Limiting reagent: potassium bromide

Explanation:

The question states that the total amount of products produced is $0.016 \:moles$ . This means that each product has $0.008\:moles$ (because of 1:1 ratio). To solve this question, we need to check each answer choice to see which one matches the given information. Let’s start with the correct answer first.

Molecule A is sodium chloride and molecule B is potassium bromide. The molecular weight of sodium chloride is $58.44\frac{g}{mol}$ and the MW of potassium bromide is $119.002\frac{g}{mol}$ . The question states that we start with of each molecule; therefore, the number of moles of each molecule is

moles of molecule A = $1g\left(\frac{1\: mol}{58.44\: g}\right) = 0.017\:mol$

moles of molecule B = $1g\left(\frac{1\: mol}{119.002g}\right) = 0.008\:mol$

Since it has the lower amount of moles, molecule B is the limiting reagent. We can now calculate the moles of products produced.

moles of product C = $0.008\: mol\:B\left(\frac{1\: mol\: C}{1\: mol\: B}\right) = 0.008\:mol\:C$

moles of product D = $0.008\: mol\:B\left(\frac{1\: mol\: D}{1\: mol\: B}\right) = 0.008\:mol\:D$

The calculated amounts matches with the information given in the question; therefore, molecule A is sodium chloride, molecule B is potassium bromide, and the limiting reagent is potassium bromide.

If we follow this procedure for the other answer choices we will notice that the results don’t match with the given information.

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Example Question #23 : Reaction Calculations And Limiting Reagent

A researcher runs an acid-base reaction using 3g of hydrochloric acid and 5g of cesium hydroxide. He wants to use the excess cesium hydroxide for another reaction. How much excess cesium hydroxide can he obtain after the completion of the reaction?

Possible Answers:

0.8g

0.2g

0.4g

Correct answer:

Explanation:

The first step is to determine the limiting reagent. To do this we need to first calculate the moles of reactants. The MW of hydrochloric acid is $36.46\frac{g}{mol}$ and MW of cesium hydroxide is $149.9\frac{g}{mol}$ . The moles of each reactant is

$3g\:HCl \left(\frac{1\: mol}{36.46g}\right) = 0.082\:mol\:HCl$

$5g\:CsOH \left(\frac{1\: mol}{149.9g}\right) = 0.033\:mol\:CsOH$

Next step is to write out the balanced chemical reaction

$HCl + CsOH \rightleftharpoons CsCl + H_2O$

The ratio of reactants is 1:1; therefore, since CsOH has the smaller amount of moles it is the limiting reagent. This means that all of cesium hydroxide will be utilized in this reaction and the researcher won’t be able to salvage any cesium hydroxide. Note that there will be excess HCl left after completion of reaction. $0.033\:moles$ of HCl will react with $0.033\:moles$ of CsOH ; therefore, there will be a total of $0.049\:moles$ of excess HCl $(0.082 - 0.033 = 0.049\:moles)$ .

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MCAT Physical : Reaction Calculations and Limiting Reagent

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #21 : Reaction Calculations And Limiting Reagent

Example Question #22 : Reaction Calculations And Limiting Reagent

Example Question #23 : Reaction Calculations And Limiting Reagent

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