Given the reaction between ammonia and oxygen as 4NH3 + 5O2 -> 4NO + 6H2O
a. Calculate the amount of nitrogen monoxide (in grams) produced if 0.8 g of ammonia is reacted.
b. Calculate the amount of nitrogen monoxide (in grams) produced if 0.8 g of oxygen is reacted.
c. In comparing (a) and (b), indicate what the limiting reactant is. Explain.
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a. The molar mass of ammonia (NH3) is approximately 17 g/mol. Therefore, 0.8 g of ammonia is 0.8/17 = 0.047 mol. According to the balanced equation, 4 mol of ammonia produces 4 mol of nitrogen monoxide (NO). Therefore, 0.047 mol of ammonia will produce 0.047 mol of NO. The molar mass of NO is approximately 30 g/mol, so 0.047 mol of NO is 0.047*30 = 1.41 g.
b. The molar mass of oxygen (O2) is approximately 32 g/mol. Therefore, 0.8 g of oxygen is 0.8/32 = 0.025 mol. According to the balanced equation, 5 mol of oxygen produces 4 mol of NO. Therefore, 0.025 mol of oxygen will produce (4/5)*0.025 = 0.02 mol of NO. The molar mass of NO is approximately 30 g/mol, so 0.02 mol of NO is 0.02*30 = 0.6 g.
c. Comparing (a) and (b), we can see that less NO is produced when the same mass of oxygen is reacted compared to when the same mass of ammonia is reacted. Therefore, oxygen is the limiting reactant because it limits the amount of NO that can be produced.
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