grams of O2? c.If we start with 8.0 moles of NH3 and 8.0 moles of O2, how many moles of NO will we produce? d. If we have 36.00 g of NH3 and 72.00 g of O2, what is the theoretical yield
An intermediate step in the industrial production of nitric acid, HNO3, involves the reaction of ammonia, NH3, with oxygen gas, O2, to form nitrogen monoxide, NO, and water, H2O.
4 NH3 (g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)
Molar masses (g mol-1) NH3 = 17.03 O2 = 32.00 NO = 30.01 H2O = 18.02
(a) Assuming excess O2, how many grams of nitrogen monoxide, NO, (i.e., the theoretical yield of NO) can be formed by the reaction of 46.00 g of ammonia, NH3
b. Assuming excess NH3, what is the theoretical yield of NO if we start with 73.00 grams of O2?
c.If we start with 8.0 moles of NH3 and 8.0 moles of O2, how many moles of NO will we produce?
d. If we have 36.00 g of NH3 and 72.00 g of O2, what is the theoretical yield of NO
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