Gas Stoichiometry At STP

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Presentation transcript:

Gas Stoichiometry At STP Start with a balanced equation. Convert mass to moles (use molar mass) or volume to moles (use 22.4 L/mol), if necessary. Use mole ratios to calculate moles of unknown. Convert moles to mass (use molar mass) or volume (use 22.4 L/mol), if necessary.

Gas Stoichiometry If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H2(g) + N2(g)  2NH3(g) 3 moles H2 + 1 mole N2  2 moles NH3 3 liters H2 + 1 liter N2  2 liters NH3

Volume-Volume Calculation How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen? 3 H2(g) + N2(g)  2NH3(g) 12 L H2 2 L NH3 = L NH3 8.0 3 L H2 Volume ratio

Mass-Volume Calculation How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) Molar volume 50.0 g KClO3 1 mol KClO3 3 mol O2 22.4 L O2 122.55 g KClO3 2 mol KClO3 1 mol O2 = 13.7 L O2 Molar mass Mole ratio

Gas Stoichiometry NOT at STP Start with a balanced equation. Convert to moles (if amount is L, use n = PV/RT). Use mole ratios to calculate moles of unknown. Convert moles to desired measurement (to convert to L use PV=nRT).

Mass-Volume NOT at STP How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO3(s)  2 KCl(s) + 3 O2(g) 50.0 g KClO3 1 mol KClO3 3 mol O2 0.612 = mol O2 122.55 g KClO3 2 mol KClO3 = 16.7 L