consider the balanced equation.\nn2 + 3h2 → 2nh3\nwhat is the percent yield of nh3 if the reaction of 26.3 g…

consider the balanced equation.\nn2 + 3h2 → 2nh3\nwhat is the percent yield of nh3 if the reaction of 26.3 g of h2 produces 79.0 g of nh3?\nuse %yield = \\frac{actual yield}{theoretical yield}×100.\n17.8%\n33.3%\n35.7%\n53.4%

consider the balanced equation.\nn2 + 3h2 → 2nh3\nwhat is the percent yield of nh3 if the reaction of 26.3 g of h2 produces 79.0 g of nh3?\nuse %yield = \\frac{actual yield}{theoretical yield}×100.\n17.8%\n33.3%\n35.7%\n53.4%

Answer

Explanation:

Step1: Calculate moles of H2

Molar mass of H2 is 2 g/mol. Moles of H2 = $\frac{26.3\ g}{2\ g/mol}=13.15\ mol$.

Step2: Determine moles of NH3 from stoichiometry

From the equation $N_2 + 3H_2\rightarrow2NH_3$, mole - ratio of H2 to NH3 is 3:2. So moles of NH3 produced theoretically = $\frac{2}{3}\times13.15\ mol\approx8.77\ mol$.

Step3: Calculate theoretical mass of NH3

Molar mass of NH3 is 17 g/mol. Theoretical mass of NH3 = $8.77\ mol\times17\ g/mol = 149.09\ g$.

Step4: Calculate percent - yield

Percent - yield = $\frac{79.0\ g}{149.09\ g}\times100%\approx53.0%$.

Answer:

The closest answer is 53.4%, so the answer is D. 53.4%