given the following chemical reaction of the synthesis of dihydrogen monoxide. calculate the percent yield…
given the following chemical reaction of the synthesis of dihydrogen monoxide. calculate the percent yield of dihydrogen monoxide if 65.1 grams of hydrogen gas reacts with excess oxygen, and experimentally we get 18.1 g of product. 2 h2 + o2 -> 2 h2o note: calculate theoretical mass first, then percent yield. percents should go to the hundredths place and use a space between number and unit.
Answer
Explanation:
Step1: Calculate moles of hydrogen gas
The molar mass of $H_2$ is $M_{H_2}=2\times1.008\ g/mol = 2.016\ g/mol$. The number of moles of $H_2$, $n_{H_2}=\frac{m_{H_2}}{M_{H_2}}$, where $m_{H_2} = 65.1\ g$. So $n_{H_2}=\frac{65.1\ g}{2.016\ g/mol}\approx32.39\ mol$.
Step2: Determine moles of water produced theoretically
From the balanced - chemical equation $2H_2+O_2\rightarrow2H_2O$, the mole ratio of $H_2$ to $H_2O$ is $1:1$. So the number of moles of $H_2O$ produced theoretically, $n_{H_2O - theoretical}=n_{H_2}\approx32.39\ mol$.
Step3: Calculate theoretical mass of water
The molar mass of $H_2O$ is $M_{H_2O}=2\times1.008 + 16.00\ g/mol=18.016\ g/mol$. The theoretical mass of $H_2O$, $m_{H_2O - theoretical}=n_{H_2O - theoretical}\times M_{H_2O}=32.39\ mol\times18.016\ g/mol\approx583.54\ g$.
Step4: Calculate percent - yield
The percent - yield formula is $\text{Percent Yield}=\frac{m_{H_2O - experimental}}{m_{H_2O - theoretical}}\times100%$. Given $m_{H_2O - experimental}=18.1\ g$ and $m_{H_2O - theoretical}\approx583.54\ g$. So $\text{Percent Yield}=\frac{18.1\ g}{583.54\ g}\times100%\approx3.10%$.
Answer:
3.10 %