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Transcript

Determine volume ratios for gaseous reactants and products by using coefficients from chemical equations. Apply gas laws to calculate amounts of gaseous reactants and products in a chemical reaction. • When gases react, the coefficients in the balanced chemical equation represent both molar amounts and relative volumes. Avogadro’s Law: Equal volumes of gases contain equal numbers of moles. This means that one mole of any gas will contain the same number of moles as one mole of any other gas. Because of this, the coefficients in balanced equations not only represent mole ratios, but also volume ratios of gases (and ONLY gases.) • Example: 2H2(g) 2 mol H2 gas 2 L H2 gas → 2H2O(g) means 1 mol O2 g 2 mol water vapor and also 1 L O2 gas 2 L water vapor + O2(g) • How many liters of hydrogen gas are required to react with 1.50 L oxygen gas? • 2H2(g) + → O2(g) 1.50 𝐿 𝑂2 x 2𝐿 𝐻2 1𝐿 𝑂2 2H2O(g) = 3.00 𝐿 𝑂2 • You can use the Ideal Gas Law to derive that at STP, 1 mol of any gas has a volume of 22.4 L. In fact, give it a shot: • If P = 1 atm, T = 273 K, and n = 1 mol, what is V? • PV = nRT • (1 atm)(273 K) = (1 mol)(0.0821 • V = 22.4 L 𝐿 𝑎𝑡𝑚 )V 𝑚𝑜𝑙 𝐾 • So for gases (and gases only!), the following conversion factor can be used: 1 mol = 22.4 L N2 (g) + 3 H2 (g) → 2 NH3 (g) • Find the mass of ammonia produced if 5.00 L of nitrogen gas reacts with hydrogen gas at STP. 1 𝑚𝑜𝑙 𝑁2 2 𝑚𝑜𝑙 𝑁𝐻3 17.04 𝑔 𝑁𝐻3 5.00 𝐿 𝑁2 𝑥 𝑥 𝑥 = 7.61 𝑔 𝑁𝐻3 22.4 𝐿 𝑁2 1 𝑚𝑜𝑙 𝑁2 1 𝑚𝑜𝑙 𝑁𝐻3 N2 (g) + 3 H2 (g) → 2 NH3 (g) • Find the mass of ammonia produced if 5.00 L of nitrogen gas reacts with excess hydrogen gas at 3.00 atm and 25 oC. (Hint: you have to do the stoichiometry AND use the Ideal Gas Law because this is not at STP. The important thing is that you need to moles of nitrogen!) • V = 5.00 L P = 3.00 atm T = 25oC n=? • PV=nRT • (3.00 𝐿 𝑎𝑡𝑚 atm)(5.00L)=n(0.0821 )(298 𝑚𝑜𝑙 𝐾 • n = 0.61 mol N2 K) • 0.61 𝑚𝑜𝑙 𝑁2 𝑥 2 𝑚𝑜𝑙 𝑁𝐻3 1 𝑚𝑜𝑙 𝑁2 17.04 𝑔𝑁𝐻3 𝑥 1 𝑚𝑜𝑙 𝑁𝐻3 = 21 𝑔 𝑁𝐻3