Limiting reagents with lots of conversions?
4 NH3 (g) + 5 O2(g) --> 4 NO(g) + 6 H2O(g)
in a certain experiment, 1.50 g of NH3 reacts with 2.75 g of O2.
which is the limiting reagent?
how many grams of NO and of H2O form?
how many grams of the excess reagent remain after the limiting reactant is completely consumed?
how are those calculations consistent with the law of conservation of mass?
Answer:
Molar mass NH3 = 17 g/mol
1.50 / 17 = 0.0882 moles NH3
Molar mass O2 = 32 g/mol
2.75 / 32 = 0.0859 moles O2
the ratio between NH3 and N2 is 4 : 5
4 : 5 = x : 0.0859
x = 0.0687 moles NH3 needed for the reaction so O2 is the limiting reactant
The ratio between O2 (limiting reactant ) and NO is 5 :4
5 : 4 = 0.0859 : x
x = 0.0687 moles NO produced
molar mass NO = 30 g/mol
0.0687 mol x 30 g/mol = 2.06 g NO produced
the ratio between O2 and H2O is 5 : 6
5 : 6 = 0.0859 : x
x = 0.103 moles H2O produced
molar mass H2O = 18 g/mol
0.103 x 18 = 1.85 g H2O produced
4 : 5 = x : 0.0859
x = 0.0687 moles NH3 needed for the reaction
moles NH3 in excess = 0.0882 - 0.0687 = 0.0195 moles NH3 in excess
0.0195 x 17 = 0.331 g of NH3 in excess
Total grams of reactants = 1.50 + 2.75 = 4.25 g
total grams of products = 2.06 + 1.85 = 3.91 g
but we have 0.331 of NH3 (reactant in excess )
so at last we have 4.25 - 0.331 = 3.91 g of reactants >> 3.91 g of products
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