what is the maximum mass of s₈ that can be produced by combining 75.0 g of each reactant? 8so₂ + 16h₂s → 3s₈…

Explanation:

Step1: Calculate molar masses

The molar mass of $SO_{2}$: $M_{SO_{2}}=32.07 + 2\times16.00=64.07\ g/mol$. The molar mass of $H_{2}S$: $M_{H_{2}S}=2\times1.01+32.07 = 34.09\ g/mol$. The molar mass of $S_{8}$: $M_{S_{8}}=8\times32.07 = 256.56\ g/mol$.

Step2: Determine moles of reactants

The number of moles of $SO_{2}$, $n_{SO_{2}}=\frac{m_{SO_{2}}}{M_{SO_{2}}}=\frac{75.0\ g}{64.07\ g/mol}\approx1.17\ mol$. The number of moles of $H_{2}S$, $n_{H_{2}S}=\frac{m_{H_{2}S}}{M_{H_{2}S}}=\frac{75.0\ g}{34.09\ g/mol}\approx2.20\ mol$.

Step3: Use stoichiometry

From the balanced equation $8SO_{2}+16H_{2}S\rightarrow3S_{8}+16H_{2}O$, the mole - ratio of $SO_{2}$ to $H_{2}S$ is $\frac{n_{SO_{2}}}{n_{H_{2}S}}=\frac{8}{16}=\frac{1}{2}$. For the given amounts, $\frac{n_{SO_{2}}}{n_{H_{2}S}}=\frac{1.17\ mol}{2.20\ mol}\approx0.53$. The limiting reactant is determined by comparing the mole - ratio of reactants to the stoichiometric ratio. According to the equation, for every 8 moles of $SO_{2}$, 16 moles of $H_{2}S$ are required. Based on the stoichiometry, from 8 moles of $SO_{2}$, 3 moles of $S_{8}$ are produced. If $n_{SO_{2}} = 1.17\ mol$, the moles of $S_{8}$ produced, $n_{S_{8}}=\frac{3}{8}\times n_{SO_{2}}=\frac{3}{8}\times1.17\ mol\approx0.439\ mol$.

Step4: Calculate mass of $S_{8}$

The mass of $S_{8}$, $m_{S_{8}}=n_{S_{8}}\times M_{S_{8}}=0.439\ mol\times256.56\ g/mol\approx113\ g$.

Answer:

$113\ g$