How many moles of NH3 are necessary to produce 0.824 mol N2
Answer:
1.648 moles of ammonia will be required to produce 0.824 moles of nitrogen gas.
Explanation:
\(2NH_3\rightarrow N_2+3H_2\)
According to reaction 1 mole of nitrogen gas is produced by 2 moles of ammonia.
Then, 0.824 moles of nitrogen gas will be produced by:
\(\frac{2}{1}\times 0.824 mol=1.648 mol\)
1.648 moles of ammonia will be required to produce 0.824 moles of nitrogen gas.
0.824 moles of nitrogen gas will be produced by 0.412 mol \(\bold {NH_3}\) since the molar ratio of ammonia and nitrogen gas is 2:1.
A mole is the measurement of a certain number of molecules, atoms, or ions.
1 mole = \(6.02 \times 10^2^3\)
Two moles \(\bold {NH_3}\) are dissociated to produce 1 mole of \(\bold N_2 {}\) and 3 moles of \(\bold {H_2}\).
\(\bold {2NH_3 \rightarrow N_2 +3H_2}\)
Therefore, 0.824 moles of nitrogen gas will be produced by 0.412 mol \(\bold {NH_3}\), since the molar ratio of ammonia and nitrogen gas is 2:1.
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