To solve the problem of determining the amount of moles of products and remaining reactants after a chemical reaction, we will follow these steps:
1. Identify the Initial Quantities:
- Initially, we have 0.280 mol of [tex]\( O_2 \)[/tex].
- Initially, we have 0.871 mol of [tex]\( H_2 \)[/tex].
2. Determine the Limiting Reactant:
- The balanced chemical equation is:
[tex]\[
O_2(g) + 2 H_2(g) \longrightarrow 2 H_2O(l)
\][/tex]
- According to the equation, 1 mole of [tex]\( O_2 \)[/tex] reacts with 2 moles of [tex]\( H_2 \)[/tex]. Therefore, to fully react 0.280 mol of [tex]\( O_2 \)[/tex], we need:
[tex]\[
0.280 \text{ mol } O_2 \times 2 = 0.560 \text{ mol } H_2
\][/tex]
- We compare this with the available [tex]\( H_2 \)[/tex]:
[tex]\[
0.871 \text{ mol } H_2 \geq 0.560 \text{ mol } H_2
\][/tex]
- Since we have enough [tex]\( H_2 \)[/tex], [tex]\( O_2 \)[/tex] is the limiting reactant.
3. Calculate the Amount of Each Substance After the Reaction:
- All 0.280 mol of [tex]\( O_2 \)[/tex] will be consumed. Therefore:
[tex]\[
\text{Moles of } O_2 \text{ used} = 0.280 \text{ mol}
\][/tex]
- Accordingly, the amount of [tex]\( H_2 \)[/tex] used is:
[tex]\[
\text{Moles of } H_2 \text{ used} = 0.280 \text{ mol} \times 2 = 0.560 \text{ mol}
\][/tex]
- The remaining [tex]\( H_2 \)[/tex] is:
[tex]\[
0.871 \text{ mol} - 0.560 \text{ mol} = 0.311 \text{ mol}
\][/tex]
- The moles of [tex]\( H_2O \)[/tex] produced are:
[tex]\[
2 \times 0.280 \text{ mol} = 0.560 \text{ mol}
\][/tex]
4. Summarize the Results:
- Moles of water ([tex]\( H_2O \)[/tex]) produced: 0.560 mol
- Moles of remaining hydrogen ([tex]\( H_2 \)[/tex]): 0.311 mol
- Moles of remaining oxygen ([tex]\( O_2 \)[/tex]): 0.000 mol
Therefore:
[tex]\[
H_2 O : 0.560 \text{ mol}
\][/tex]
[tex]\[
H_2 : 0.311 \text{ mol}
\][/tex]
[tex]\[
O_2 : 0.000 \text{ mol}
\][/tex]
