consider the chemical equation.\n2h₂ + o₂ → 2h₂o\nwhat is the percent yield of h₂o if 87.0 g of h₂o is…

Explanation:

Step1: Determine the limiting reactant

First, find the moles of $H_2$ and $O_2$. Molar - mass of $H_2$ is $M_{H_2}=2.02\ g/mol$, molar - mass of $O_2$ is $M_{O_2}=32.00\ g/mol$. Moles of $H_2,n_{H_2}=\frac{m_{H_2}}{M_{H_2}}=\frac{11.0\ g}{2.02\ g/mol}\approx5.45\ mol$. Moles of $O_2,n_{O_2}=\frac{m_{O_2}}{M_{O_2}}=\frac{95.0\ g}{32.00\ g/mol}\approx2.97\ mol$. From the balanced equation $2H_2 + O_2\rightarrow2H_2O$, the mole - ratio of $H_2$ to $O_2$ is $2:1$. For $n_{O_2} = 2.97\ mol$, the moles of $H_2$ required is $n_{H_2\ required}=2\times n_{O_2}=2\times2.97\ mol = 5.94\ mol$. But we have only $5.45\ mol$ of $H_2$. So, $H_2$ is the limiting reactant.

Step2: Calculate the theoretical yield of $H_2O$

From the balanced equation, the mole - ratio of $H_2$ to $H_2O$ is $1:1$. Since $n_{H_2}=5.45\ mol$, the moles of $H_2O$ produced theoretically, $n_{H_2O\ theoretical}=5.45\ mol$. Molar - mass of $H_2O$ is $M_{H_2O}=18.02\ g/mol$. Theoretical yield of $H_2O,m_{H_2O\ theoretical}=n_{H_2O\ theoretical}\times M_{H_2O}=5.45\ mol\times18.02\ g/mol\approx98.2\ g$.

Step3: Calculate the percent yield

Given actual yield $m_{H_2O\ actual}=87.0\ g$. Percent yield $=\frac{m_{H_2O\ actual}}{m_{H_2O\ theoretical}}\times100=\frac{87.0\ g}{98.2\ g}\times100\approx88.6% \approx88.5%$ (rounding difference).

Answer:

88.5%