Introduction to Stoichiometry Worksheet

This worksheets consists of introductory questions to the concept of a mole, and some simple stoichiometry problems.

1. What is a mole? Easy
   Reveal answer
   A mole is the SI unit for the amount of a substance, representing a quantity of $6.022 \times 10^{23}$ atoms/molecules/ions.
2. One mole of liquid water and one mole of liquid mercury weigh the same amount. True/False? Easy
   Reveal answer
   False. Water has a far lower molar mass than mercury, thus one mole of $H_2O$ will weigh far less than one mole of $Hg$.
3. In a chemical equation, the coefficients before reactants represent the molar proportions in which they react. True/False? Medium
   Reveal answer
   True. While these coefficients can also be thought of as representing molecules/atoms/ions, the more general way to think about them is in terms of moles.

1. What is the molar mass of benzene $(C_6H_6)$ gas? Medium
   Reveal answer
   The molar mass of compounds is determined by summing their elemental constituents. In the case of benzene, we have $$6 \times \text{C atoms} \implies 6 \times M_C = 6 \times 12.011 = 72.066 \text{ g/mol}$.$$ $$6 \times \text{H atoms} \implies 6 \times M_H = 6 \times 1.008 = 6.048 \text{ g/mol}$.$$ Thus, the molar mass of benzene is $72.066+6.048=\mathbf{78.114} \textbf{ g/mol}$.
2. How many moles of toluene $(C_7H_8)$ are there in 12g? Medium
   Reveal answer
   The molar mass of toluene is 92.141 g/mol, by the same method as in Q1. In 12g, we have $n$ moles, where $$n = \frac{m}{M} = \frac{12}{92.141} = 0.127 \text{ mol}$$ Thus, we have 0.127 moles of toluene in 12g.
3. Imagine 30g of both propane and oxygen gas are pumped into a glass box with nothing else inside it (i.e. a vacuum). A match is then dropped into the box, somehow without letting any air in, which causes a big explosion as per the reaction below:
   
        C3H8 + 5O2 → 3CO2 + 4H2O
   
   Assume the reaction runs to completion, and following this, more oxygen is pumped in and another match thrown in. Will a second explosion occur in the box? Hard
   Reveal answer
   To answer this, we need to work out which of the reactants is the limiting reagent (i.e. is used up in the reaction), and which remains in excess after combustion is finished. The molar mass of propane gas is $3(12.011)+8(1.008)=44.097$ g/mol, whereas for oxygen gas, the molar mass is $2(16.00)=32$ g/mol. Thus, we have the following number of moles of each: $$n_{O_2}=\frac{m}{M}=\frac{30}{32}=0.9375 \text{ moles}$$ $$n_{Propane}=\frac{m}{M}=\frac{30}{44.097}=0.6803 \text{ moles}$$ but, each mole of propane reacts with 5 of oxygen, thus our 0.9375 moles of oxygen will be used up by $\frac{0.9375}{5}=0.1875$ moles of propane, leaving $0.6803-0.1875=0.4928$ moles of propane gas in the box after combustion occurs to completion. Therefore, propane is the gas remaining in the box, oxygen is the limiting reagent, and thus if more oxygen is pumped in and ignited, another explosion will indeed occur.