N2 (g) + 3H2 (g) -------------------------> 2NH3 (g)
(3.4 mol)
Considering the stoichiometry,
1.
"\\qquad\\qquad\n\\begin {aligned}\n\\frac{n_{N_2}}{n_{H_2}} &= \\frac{1}{3}\\\\\n\\\\\n\\frac{3.4}{n_{H_2}} &= \\frac{1}{3}\\\\\n\\\\\nn_{H_2} &= \\bold{10.2\\, mol}\n\n\n\\end {aligned}"
2.
"\\qquad\\qquad\n\\begin {aligned}\n\\frac{n_{N_2}}{n_{NH_3}} &= \\frac{1}{2}\\\\\n\\\\\n\\frac{3.4}{n_{NH_3}} &= \\frac{1}{2}\\\\\n\\\\\nn_{NH_3} &= \\bold{6.8\\, mol}\n\\end{aligned}"
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