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Hydrogen gas and nitrogen gas react in the presence of a catalyst to produce ammonia gas. rnrncalculate the %yield in an experiment where 144.0g of hydrogen gas yielded 436g of ammonia gas experimentally.
Accepted Answer
The balanced chemical equation for the reaction is:
N2 + 3H2 → 2NH3
From the equation, we see that 3 moles of hydrogen react with 1 mole of nitrogen to produce 2 moles of ammonia. The molar mass of hydrogen is 2.02 g/mol, and the molar mass of ammonia is 17.04 g/mol.
Using the given mass of hydrogen and the molar masses, we can calculate the number of moles of hydrogen used:
144.0 g H2 * (1 mol H2 / 2.02 g H2) = 71.3 mol H2
According to the balanced equation, 71.3 mol of hydrogen should react with 23.8 mol of nitrogen to produce 47.6 mol of ammonia.
The experimental yield is 436 g of ammonia, which can be converted to moles using the molar mass of ammonia:
436 g NH3 * (1 mol NH3 / 17.04 g NH3) = 25.6 mol NH3
The percent yield is calculated as follows:
% yield = (experimental yield / theoretical yield) * 100%
% yield = (25.6 mol NH3 / 47.6 mol NH3) * 100%
% yield = 53.8%
Therefore, the percent yield in this experiment is 53.8%.
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