Question 8.8: What is the volume, in liters, of 64.0 g of O2 gas at STP?
What is the volume, in liters, of 64.0 g of O_{2} gas at STP?
STEP 1 State the given and needed quantities (moles).
| ANALYZE THE PROBLEM | Given | Need | Connect |
| 64.0 g of O_{2}(g) at STP | liters of O_{2} gas at STP | molar mass, molar volume (STP) |
STEP 2 Write a plan to calculate the needed quantity.
grams of O_{2} \boxed{Molar Mass} → moles of O_{2} \boxed{Molar volume} → liters of O_{2}STEP 3 Write the equalities and conversion factors including 22.4 L/mole at STP.
\begin{array}{r c}\boxed{\begin{matrix} 1 mole of O_{2} = 32.00 g of O_{2} \\\frac{32.00 g O_{2}}{1 mole O_{2}} \text{ and } \frac{1 mole O_{2}}{32.00 g O_{2}} \end{matrix}} & \boxed{\begin{matrix} 1 mole of O_{2} = 22.4 L of O_{2}(STP) \\ \frac{22.4 L O_{2}(STP)}{ 1 mole O_{2}}\text{ and }\frac{1 mole O_{2}}{22.4 L O_{2}(STP)} \end{matrix}}\end{array}STEP 4 Set up the problem with factors to cancel units.
64.0 \cancel{g O_{2}} \times \boxed{\frac{1 \cancel{mole O_{2}}}{32.00 \cancel{ g O_{2}}} } \times \boxed{\frac{22.4 L O_{2} (STP)}{1 \cancel{ mole O_{2}}} } = 44.8 L of O_{2} (STP)Related Answered Questions
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