2 C2H6(g) + 2 O2(g) = CO2(g) + 2 H2O(I)
1. How many grams of H2O will be produced by 58.2 L of CH4 at STP? Assume an excess of O2.
We use the fact that 1 mole of any gas at STP has a volume of 22.4 L to find how much water was produced:
"\\text{58.2 L of CH}_4 \\text{ at STP} \\times\\cfrac{\\text{1 mol of CH}_4}{\\text{22.4 L of CH}_4}\\times\\cfrac{\\text{2 mol of H}_2O}{\\text{1 mol of CH}_4}\\times\\cfrac{\\text{18 g of H}_2O}{\\text{1 mol of H}_2O}\n\\\\ =93.54\\text{ g of H}_2O"
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