Question

question 1
so₂ reacts with h₂s as follows:
2h₂s + so₂ ⇌ 3s + 2h₂o
when 7.50 g of h₂s reacts with 12.75 g of so₂, which statement applies?
so₂ is the limiting reagent.
1.13 g of h₂s remain.
10.6 g of sulfur are formed.
6.38 g of sulfur are formed.
0.0216 moles of h₂s remain.

Explanation:

Step1: Calculate moles of reactants

The molar mass of $H_2S$ is $M_{H_2S}=2\times1 + 32=34\ g/mol$. The number of moles of $H_2S$, $n_{H_2S}=\frac{7.50\ g}{34\ g/mol}\approx0.2206\ mol$. The molar mass of $SO_2$ is $M_{SO_2}=32 + 2\times16 = 64\ g/mol$. The number of moles of $SO_2$, $n_{SO_2}=\frac{12.75\ g}{64\ g/mol}\approx0.1992\ mol$.

Step2: Determine the limiting - reagent

From the balanced equation $2H_2S+SO_2
ightarrow3S + 2H_2O$, the mole - ratio of $H_2S$ to $SO_2$ is $2:1$. For $n_{SO_2} = 0.1992\ mol$, the amount of $H_2S$ required for complete reaction is $n_{H_2S\ required}=2\times n_{SO_2}=2\times0.1992\ mol = 0.3984\ mol$. Since we have only $0.2206\ mol$ of $H_2S$, $H_2S$ is the limiting reagent.

Step3: Calculate moles of $SO_2$ reacted

The moles of $SO_2$ that react with $0.2206\ mol$ of $H_2S$ is $n_{SO_2\ reacted}=\frac{0.2206\ mol}{2}=0.1103\ mol$.

Step4: Calculate moles of $H_2S$ remaining

The moles of $H_2S$ remaining is $n_{H_2S\ remaining}=0$ (since $H_2S$ is the limiting reagent and is completely consumed in the reaction).

Step5: Calculate moles of sulfur formed

From the balanced equation, for every 2 moles of $H_2S$ reacted, 3 moles of $S$ are formed. So, the moles of $S$ formed from $0.2206\ mol$ of $H_2S$ is $n_S=\frac{3}{2}\times0.2206\ mol = 0.3309\ mol$. The molar mass of $S$ is $32\ g/mol$, so the mass of $S$ formed is $m_S=0.3309\ mol\times32\ g/mol = 10.5888\ g\approx10.6\ g$.

Answer:

10.6 g of sulfur are formed.