Ammonia gas is formed from nitrogen gas and hydrogen gas according to the following equation: N2 (g) + 3H2 (g) ➡ 2NH3 (g). If 112 grams of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 120 grams of ammonia, what is the percent yield of this reaction?

44.1%

66.2%

88.3%

96.4%

Question

Ammonia gas is formed from nitrogen gas and hydrogen gas according to the following equation: N2 (g) + 3H2 (g) ➡  2NH3 (g). If 112 grams of nitrogen gas is allowed to react with an excess of hydrogen gas to produce 120 grams of ammonia, what is the percent yield of this reaction?
		
44.1%
		
66.2%
		
88.3%
		
96.4%

rhondasommer · Answer

@sweetburger

rhondasommer · Answer

this one is 44.1? I think?

sweetburger · Answer

YOu have 4 moles of N2 which should theoretically produce 8 moles of NH3. However, the reaction only produces roughly 7 moles or 7.06 moles of NH3 so 7.06/8 x 100% = 88.3%

rhondasommer · Answer

but why does 4moles of N2 = 8moles of NH3?

sweetburger · Answer

1:2 ratio between N2 to NH3. For every 1 mole of N2 you make 2 moles of NH3. so 4 moles of N2 makes 8 moles NH3

rhondasommer · Answer

ohhhh okay I get it ... i think lol