Question

question 19 using the equation 2li + 2h₂o → 2lioh + h₂ when 2.5 g of li reacts with water 0.34 g of h₂ is produced. the percent yield of the reaction is blank 1% blank 1 add your answer 5 points listen

Explanation:

Step1: Calculate the theoretical yield of H₂

First, find the molar mass of Li. The molar mass of Li ($M_{Li}$) is approximately 6.94 g/mol. The balanced chemical equation is $2Li + 2H_2O
ightarrow2LiOH + H_2$. From the equation, 2 moles of Li produce 1 mole of $H_2$. The number of moles of Li ($n_{Li}$) is $n_{Li}=\frac{m_{Li}}{M_{Li}}=\frac{2.5g}{6.94g/mol}\approx0.3602mol$. According to the stoichiometry, the number of moles of $H_2$ produced theoretically ($n_{H_2 - theoretical}$) is $\frac{1}{2}n_{Li}=\frac{1}{2}\times0.3602mol = 0.1801mol$. The molar mass of $H_2$ ($M_{H_2}$) is 2.02 g/mol. So the theoretical mass of $H_2$ ($m_{H_2 - theoretical}$) is $m_{H_2 - theoretical}=n_{H_2 - theoretical}\times M_{H_2}=0.1801mol\times2.02g/mol\approx0.3638g$.

Step2: Calculate the percent - yield

The percent - yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$. The actual yield of $H_2$ is $m_{H_2 - actual}=0.34g$. So the percent - yield is $\text{Percent Yield}=\frac{0.34g}{0.3638g}\times100\%\approx93.5\%$.

Answer:

93.5%