Question

1H2 + 1F2 = 2HF
products: 8HF
leftovers: 3H2, 0F2
what are the reactants?

Answer 1

To determine the reactants, we need to analyze the given information.

The equation provided is:

\[
\text{1H}_2 + \text{1F}_2 \rightarrow 2\text{HF}
\]

This indicates that one mole of hydrogen gas \((\text{H}_2)\) reacts with one mole of fluorine gas \((\text{F}_2)\) to produce two moles of hydrogen fluoride \((\text{HF})\).

You mentioned that there are 8 moles of HF produced as products. Since the balanced chemical equation shows that 2 moles of HF are produced from 1 mole of \(\text{H}_2\) and 1 mole of \(\text{F}_2\), the following stoichiometric relationship applies:

- To produce 8 moles of HF, you would need:

\[
\frac{8 \text{ HF}}{2 \text{ HF per reaction}} = 4 \text{ reactions}
\]

For these 4 reactions, you would require:

- 4 moles of \(\text{H}_2\)
- 4 moles of \(\text{F}_2\)

However, you also mentioned that you were left with 3 moles of \(\text{H}_2\) and no \(\text{F}_2\) remaining. This means that out of the 4 moles of \(\text{H}_2\) you initially had, 1 mole of \(\text{H}_2\) reacted (since 4 - 3 = 1).

Since you have 0 moles of \(\text{F}_2\) remaining, it indicates that all 4 moles of \(\text{F}_2\) were consumed in the reactions.

### Summary of the reactants:
- Initial reactants: 4 moles of \(\text{H}_2\) and 4 moles of \(\text{F}_2\)
- Reacted: 1 mole of \(\text{H}_2\) and 4 moles of \(\text{F}_2\)

Thus, the starting reactants were:
- \(4 \text{H}_2\)
- \(4 \text{F}_2\)