Question

question 5 (2 points)
what is the theoretical yield in grams of hydrogen gas when 6.54 g zn reacts with 7.96 g of h₂so₄? report your answer to 3 decimal places but do not include the unit of grams in your response.
zn (s) + h₂so₄ (aq) → h₂ (g) + znso₄

Explanation:

Step1: Calculate moles of Zn

The molar - mass of Zn is $M_{Zn}=65.38\ g/mol$. The number of moles of Zn, $n_{Zn}=\frac{m_{Zn}}{M_{Zn}}=\frac{6.54\ g}{65.38\ g/mol}\approx0.1\ mol$.

Step2: Calculate moles of $H_2SO_4$

The molar - mass of $H_2SO_4$ is $M_{H_2SO_4}=(2\times1 + 32+4\times16)\ g/mol = 98\ g/mol$. The number of moles of $H_2SO_4$, $n_{H_2SO_4}=\frac{m_{H_2SO_4}}{M_{H_2SO_4}}=\frac{7.96\ g}{98\ g/mol}\approx0.0812\ mol$.

Step3: Determine the limiting reactant

From the balanced chemical equation $Zn(s)+H_2SO_4(aq)
ightarrow H_2(g)+ZnSO_4(aq)$, the mole ratio of $Zn$ to $H_2SO_4$ is 1:1. Since $n_{H_2SO_4}<n_{Zn}$, $H_2SO_4$ is the limiting reactant.

Step4: Calculate moles of $H_2$ produced

The mole ratio of $H_2SO_4$ to $H_2$ is 1:1. So, $n_{H_2}=n_{H_2SO_4}\approx0.0812\ mol$.

Step5: Calculate the mass of $H_2$ produced

The molar - mass of $H_2$ is $M_{H_2}=2\ g/mol$. The mass of $H_2$, $m_{H_2}=n_{H_2}\times M_{H_2}=0.0812\ mol\times2\ g/mol = 0.162\ g$.

Answer:

0.162