Question 5.13: How many grams of water form when 1.24 L of H2 gas at STP co...
How many grams of water form when 1.24 L of H_{2} gas at STP completely reacts with O_{2}?
2 H_{2}(g) + O_{2}(g) → 2 H_{2}O(g)
| SORT You are given the volume of hydrogen gas (a reactant) at STP and asked to determine the mass of water that forms upon complete reaction. |
GIVEN 1.24 L H_{2} FIND g H_{2}O |
| STRATEGIZE Since the reaction occurs under standard temperature and pressure, you can convert directly from the volume (in L) of hydrogen gas to the amount in moles. Then use the stoichiometric relationship from the balanced equation to find the number of moles of water that forms. Finally, use the molar mass of water to obtain the mass of water. | CONCEPTUAL PLAN L H2 → mol H2 → mol H2O → g H2O . \frac{1 mol H_{2}}{22.4 L H_{2}} \frac{2 mol H_{2}O}{2 mol H_{2}} \frac{18.02 g}{1 mol} RELATIONSHIPS USED |
| SOLVE Follow the conceptual plan to solve the problem. | |
1.24\cancel{ L H_{2}} \times\frac{1 \cancel{mol H_{2}}}{22.4 \cancel{L H_{2}}}\times\frac{2 \cancel{mol H_{2}O}}{2 \cancel{mol H_{2}}}\times \frac{18.02 g H_{2}O}{1 \cancel{mol H_{2}O}}
= 0.998 g H_{2}O
CHECK The units of the answer are correct. The magnitude of the answer (0.998 g) is about 1/18 of the molar mass of water, roughly equivalent to the approximately 1/22 of a mole of hydrogen gas given, as expected for a 1:1 stoichiometric relation-ship between number of moles of hydrogen and number of moles of water.