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2.00 g sample of ammonia is mixed with 4.00 g of oxygen. How much excess reactant remains after the reaction has stopped?
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Anonymous
molar mass of NH3 = 14 + 3 = 17 grams/mol
so you have 2 grams ( 1 mol/17 g) = 0.118 mols of NH3
molar mass of O2 = 16*2 = 32 g /mol
so you have 4 g (1 mol/32 g) = 1/8 = 0.125 mols of O2
NH3 + O2 ----> H2O + N2 maybe? If N is not oxidized to NO2 or worse
Balance that
4 NH3 + 3 O2 ---> 6H2O + 2 N2
so for .118 mols of NH3 you need (3/4) mols of O2 = 0.0885 mols of O2
.0885 mols * 32 g/mol = 2.832 grams of O2 needed
but we have 4 grams of O2
so how much O2 is left over?
so you have 2 grams ( 1 mol/17 g) = 0.118 mols of NH3
molar mass of O2 = 16*2 = 32 g /mol
so you have 4 g (1 mol/32 g) = 1/8 = 0.125 mols of O2
NH3 + O2 ----> H2O + N2 maybe? If N is not oxidized to NO2 or worse
Balance that
4 NH3 + 3 O2 ---> 6H2O + 2 N2
so for .118 mols of NH3 you need (3/4) mols of O2 = 0.0885 mols of O2
.0885 mols * 32 g/mol = 2.832 grams of O2 needed
but we have 4 grams of O2
so how much O2 is left over?
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