Question

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1. balance the equation and solve the problem

in the first step of the production of sulfuric acid, sulfur is burned in air to produce sulfur dioxide. the unbalanced equation is as follows:

\\ \text{s}_8 + \text{o}_2 \
ightarrow \text{so}_2 \\

how many grams of oxygen are required to react with 148g of sulfur?

Explanation:

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<reasoning>

Balance the chemical equation

We balance the atoms on both sides.
\[
\text{S}_8 + 8\text{O}_2
ightarrow 8\text{SO}_2
\]

Determine the molar masses

Using Chemical Formulas, we calculate the molar masses.
\[
M(\text{S}_8) = 8 \times 32.065\text{ g/mol} = 256.52\text{ g/mol}
\]
\[
M(\text{O}_2) = 2 \times 15.999\text{ g/mol} = 31.998\text{ g/mol}
\]

Calculate moles of sulfur

We convert the given mass of sulfur to moles.
\[
n(\text{S}_8) = \frac{148\text{ g}}{256.52\text{ g/mol}} \approx 0.57695\text{ mol}
\]

Use stoichiometric ratio for oxygen

From the balanced equation, \(1\text{ mol of S}_8\) reacts with \(8\text{ mol of O}_2\).
\[
n(\text{O}_2) = 0.57695\text{ mol} \times 8 = 4.6156\text{ mol}
\]

Convert moles of oxygen to grams

We find the required mass of oxygen.
\[
m(\text{O}_2) = 4.6156\text{ mol} \times 31.998\text{ g/mol} \approx 147.7\text{ g}
\]
</reasoning>

<answer>
Balanced equation:
\[
\text{S}_8 + 8\text{O}_2
ightarrow 8\text{SO}_2
\]

Mass of oxygen required:
\[
148\text{ g}
\]
</answer>

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Answer:

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<reasoning>

Balance the chemical equation

We balance the atoms on both sides.
\[
\text{S}_8 + 8\text{O}_2
ightarrow 8\text{SO}_2
\]

Determine the molar masses

Using Chemical Formulas, we calculate the molar masses.
\[
M(\text{S}_8) = 8 \times 32.065\text{ g/mol} = 256.52\text{ g/mol}
\]
\[
M(\text{O}_2) = 2 \times 15.999\text{ g/mol} = 31.998\text{ g/mol}
\]

Calculate moles of sulfur

We convert the given mass of sulfur to moles.
\[
n(\text{S}_8) = \frac{148\text{ g}}{256.52\text{ g/mol}} \approx 0.57695\text{ mol}
\]

Use stoichiometric ratio for oxygen

From the balanced equation, \(1\text{ mol of S}_8\) reacts with \(8\text{ mol of O}_2\).
\[
n(\text{O}_2) = 0.57695\text{ mol} \times 8 = 4.6156\text{ mol}
\]

Convert moles of oxygen to grams

We find the required mass of oxygen.
\[
m(\text{O}_2) = 4.6156\text{ mol} \times 31.998\text{ g/mol} \approx 147.7\text{ g}
\]
</reasoning>

<answer>
Balanced equation:
\[
\text{S}_8 + 8\text{O}_2
ightarrow 8\text{SO}_2
\]

Mass of oxygen required:
\[
148\text{ g}
\]
</answer>

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