First, relate moles:
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Question & Answer: A student runs the reaction below starting with 5.8 grams of ammonia (NH3) and 6.0 grams of…..
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4 mol of NH3 required = 5 mol of O2
get moles of NH3:
mol = mass/MW = 5.8/17.04 = 0.340375 mol of NH3
get moles of O2:
mol = mass/MW = 6/32 = 0.1875 mol of O2
clearly, O2 is limiting
then,
5 mol of H2O = 6 mol of H2O
0.1875 mol of H2O = 6/5*0.1875 = 0.225 mol of H2O
mass of water = moil*MW = 0.225*18.02 = 4.05 g of water