To determine how many moles of SnO₂ are needed to produce 500.0 grams of tin (Sn), we can follow a structured approach involving stoichiometry and molar mass relationships.
1. Identify the chemical reaction and relevant molar masses:
The provided chemical reaction is:
[tex]\[
SnO_2 + 2 H_2 \rightarrow Sn + 2 H_2O
\][/tex]
The relevant molar masses are:
- Molar mass of \(Sn\) (Tin): 118.71 g/mol
- Molar mass of \(SnO_2\) (Tin(IV) oxide): 150.71 g/mol
2. Calculate the number of moles of Tin (Sn):
We are given 500.0 grams of Tin (Sn). To find the moles of Sn, we use the formula:
[tex]\[
\text{moles of Sn} = \frac{\text{mass of Sn}}{\text{molar mass of Sn}}
\][/tex]
Substituting the given values:
[tex]\[
\text{moles of Sn} = \frac{500.0 \text{ g}}{118.71 \text{ g/mol}} \approx 4.21 \text{ moles}
\][/tex]
3. Relate moles of Sn to moles of SnO₂:
From the balanced chemical equation, we see that 1 mole of Sn is produced from 1 mole of SnO₂ (the stoichiometric coefficients are 1:1 for SnO₂ to Sn):
[tex]\[
1 \text{ mole of SnO}_2 \rightarrow 1 \text{ mole of Sn}
\][/tex]
Therefore, the moles of SnO₂ needed to produce 4.21 moles of Sn are the same as the moles of Sn:
[tex]\[
\text{moles of SnO}_2 = 4.21 \text{ moles}
\][/tex]
So, the number of moles of \( SnO_2 \) needed to produce 500.0 grams of \( Sn \) is \( 4.21 \). The correct answer is:
[tex]\[
\boxed{4.21}
\][/tex]
