To solve how many moles of \( \text{Sn} \) are required to react completely with \( 40 \) grams of HF, you should follow these steps:
1. Find the number of moles of HF:
The molar mass of HF is given as \( 20.01 \, \text{g/mol} \).
[tex]\[
\text{Number of moles of HF} = \frac{\text{mass of HF}}{\text{molar mass of HF}}
\][/tex]
[tex]\[
\text{Number of moles of HF} = \frac{40 \, \text{g}}{20.01 \, \text{g/mol}} \approx 1.999 \, \text{moles}
\][/tex]
2. Use the stoichiometry of the reaction:
The balanced chemical equation is:
[tex]\[
\text{Sn} + 2 \, \text{HF} \rightarrow \text{SnF}_2 + \text{H}_2
\][/tex]
According to this equation, 1 mole of \( \text{Sn} \) reacts with 2 moles of \( \text{HF} \).
3. Calculate the moles of \( \text{Sn} \) required:
[tex]\[
\text{Moles of Sn} = \frac{\text{Moles of HF}}{2}
\][/tex]
[tex]\[
\text{Moles of Sn} = \frac{1.999 \, \text{moles HF}}{2} = 0.9995 \, \text{moles Sn}
\][/tex]
Therefore, approximately \( 1 \) mole of \( \text{Sn} \) is required to react completely with \( 40 \) grams of HF. Hence, from the given options, the moles of \( \text{Sn} \) required is:
[tex]\[
\boxed{1 \, \text{mole}}
\][/tex]
