AP Chemistry - Stoichiometric Calculations and Dimensional Analysis

Consider the unbalanced equation for the combustion of pentane:

$\text{C}_5\text{H}_{12}(g)+\text{O}_2(g)\rightarrow \text{CO}_2(g)+\text{H}_2\text{O}(g)$

How many grams of $\text{O}_2$ is required to react with exactly of $\text{C}_5\text{H}_{12}$ .

Explanation

Start by balancing the given chemical equation:

$\text{C}_5\text{H}_{12}(g)+8\text{O}_2(g)\rightarrow 5\text{CO}_2(g)+6\text{H}_2\text{O}(g)$

Next, convert the grams of pentane into moles of pentane.

$152.3\text{g }\text{C}_5\text{H}_{12}\cdot\frac{\text{1 molC}_5\text{H}_{12}}{72.15\text{g C}_5\text{H}_{12}}=2.110 \text{ moles of C}_5\text{H}_{12}$

Use the stoichiometric ratio given by the balanced equation to find the number of moles of oxygen needed to react with the pentane.

$2.110\text{ moles of C}_5\text{H}_{12}\cdot\frac{8\text{ moles of O}_2}{1\text{mole of C}_5\text{H}_{12}}=16.89\text{ moles of O}_2$

Finally, convert the number of moles of oxygen into grams of oxygen.

$16.89\text{ moles of O}_2\cdot \frac{32.00\text{g of O}_2}{1\text{ mole of O}_2}=540.39\text{g of O}_2$

There should be significant figures, so the answer is $540.4\text{g of O}_2$ .

Consider the following unbalanced reaction:

$\text{K}_2\text{SO}_4(aq)+\text{BaCl}_2(aq)\rightarrow \text{KCl}(aq)+\text{BaSO}_4(s)$

How many grams of $\text{BaSO}_4$ can be produced if $15.0\text{mL}$ of $1.20\text{M}\text{ K}_2\text{SO}_4$ is mixed with $25.0\text{mL}$ of $0.75\text{M BaCl}_2$ ?

Explanation

Start by balancing the equation:

$\text{K}_2\text{SO}_4(aq)+\text{BaCl}_2(aq)\rightarrow 2\text{KCl}(aq)+\text{BaSO}_4(s)$

Next, figure out which reactant is the limiting reactant.

$15.0\times10^{-3}\text{L K}_2\text{SO}_4\cdot\frac{1.20\text{ moles of K}_2\text{SO}_4}{1.00\text{L K}_2\text{SO}_4}\cdot\frac{\text{1 mole K}_2\text{SO}_4}{\text{1 mole BaSO}_4}=0.018\text{ moles of BaSO}_4$

$25.0\times10^{-3}\text{L BaCl}_2\cdot\frac{0.75\text{moles BaCl}_2}{1\text{L BaCl}_2}\cdot\frac{1\text{ mole BaCl}_2}{1\text{ mole BaSO}_4}=0.01875\text{ moles of BaSO}_4$

Since fewer moles of $\text{BaSO}_4$ is produced when $\text{K}_2\text{SO}_4$ is reacted, then $\text{K}_2\text{SO}_4$ must be the limiting reactant.

Convert the amount of moles of $\text{BaSO}_4$ into grams.

$0.018\text{ moles of BaSO}_4\cdot\frac{233.43\text{g BaSO}_4}{\text{1 mole BaSO}_4}=4.20\text{g BaSO}_4$

What is the mass in of a gas with a volume of and a density of $0.76: \frac{g}{mL}$ ?

Explanation

For this question, we're given the density and volume of a gas and we're asked to find the mass of the gas.

To answer this question, we'll need to use dimensional analysis. What this means is that we'll need to cancel out units in order to obtain the ones that we're looking for.

$2.5: L\cdot\frac{1000:mL}{1:L}\cdot \frac{0.76: g}{1: mL}\cdot \frac{1:kg}{1000:g}=1.9:kg$

$2H_{2}(g)+ O_{2}(g)\rightarrow 2H_{2}O (g)$

Use the equation shown to determine how many grams of water form when of $H_{2}(g)$ reacts completely with $O_{2}$ at .

of water

Explanation

Use the relations $1mol=22.4L H_{2}$ , molar mass $H_{2}O=18.02\frac{g}{mol}$ , and $2molH_{2}O=2molH_{2}$

$2.34L H_{2} * \frac{1mol H_{2}}{22.4L H_{2}}=.104mol H_{2} * \frac{2mol H_{2}O}{2molH_{2}}=.104molH_{2}O*\frac{18.02gH_{2}O}{1molH_{2}O}=1.88gH_{2}O$

Stoichiometric Calculations and Dimensional Analysis

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