Question 12.24: Calculating the Mass of a Reactant from the Volume of a Gase......
Calculating the Mass of a Reactant from the Volume of a Gaseous Product
Oxygen gas can be generated by heating KClO _3 to a high temperature.
2 \textrm{ KClO}_3 \textrm{(s) } \longrightarrow 2 \textrm{ KCl(s) } + 3 \textrm{ O}_2 \textrm{(g)}How much KClO _3 , in grams, is needed to generate 7.50 L of O _2 at a pressure of 1.00 atm and a temperature of 37°C?
Step 1 The given quantity is 7.50 L of O _2 , and the desired quantity is grams of KClO _3 .
7.50 L O _2 = ? g KClO _3
Step 2 This is a volume-of-gas-A to grams-of-B problem. The pathway, in terms of Figure 12.9, is
For the first unit change (volume-of-gas-A to moles-of-A) the molar volume conversion factor relationship needed, calculated using the ideal gas law, is
\frac{1 \textrm{mole O}_2}{25.4 \textrm{L O}_2}The complete dimensional analysis setup for the problem becomes
7.50 \cancel{\textrm{L O}_2} \times \frac{1 \cancel{\textrm{mole O}_2}}{25.4 \cancel{\textrm{L O}_2}} \times \frac{2 \cancel{\textrm{moles KClO}_3}}{3 \cancel{\textrm{moles O}_2}} \times \frac{122.55 \textrm{g KClO}_3}{1 \cancel{\textrm{mole KClO}_3}}volume A → moles A → moles B → grams B
Step 3 The solution, obtained from combining all the numerical factors, is
\frac{7.50 \times 1 \times 2 \times 122.55}{25.4 \times 3 \times 1 } \textrm{g KClO}_3 = 24.124016 \textrm{ g KClO}_3 (calculator answer)
= 24.1 g KClO _3 (correct answer)
