Question

if we want to double the amount of water (h2o) we make. we want to make four moles of h2o (instead of two) how many moles of h2 (hydrogen) do we need? stoichiometric calculations equation: 2 h₂(g) + o₂(g) → 2 h₂o(l) molecules: 2 molecules h₂ + 1 molecule o₂ → 2 molecules h₂o mass (amu): 4.0 amu h₂ + 32.0 amu o₂ → 36.0 amu h₂o amount (mol): 2 mol h₂ + 1 mol o₂ → 2 mol h₂o mass (g): 4.0 g h₂ + 32.0 g o₂ → 36.0 g h₂o the coefficients in the balanced equation give the ratio of moles of reactants and products 1 mole of hydrogen 2 moles of hydrogen 4 moles of hydrogen

Explanation:

Step1: Identify the mole ratio

From the balanced equation \(2 \text{ H}_2(\text{g}) + \text{ O}_2(\text{g})
ightarrow 2 \text{ H}_2\text{O}(\text{l})\), the mole ratio of \(\text{H}_2\) to \(\text{H}_2\text{O}\) is \(2:2\) (or simplified \(1:1\)).

Step2: Calculate moles of \(\text{H}_2\) for 4 moles of \(\text{H}_2\text{O}\)

Let \(x\) be the moles of \(\text{H}_2\) needed. Using the mole ratio: \(\frac{\text{moles of } \text{H}_2}{\text{moles of } \text{H}_2\text{O}}=\frac{2}{2}\). For 4 moles of \(\text{H}_2\text{O}\), we set up the proportion \(\frac{x}{4}=\frac{2}{2}\). Solving for \(x\), we get \(x = 4\) moles (since \(\frac{2 \text{ mol } \text{H}_2}{2 \text{ mol } \text{H}_2\text{O}}=\frac{x}{4 \text{ mol } \text{H}_2\text{O}}\), cross - multiplying gives \(2\times4 = 2x\), so \(x = 4\)).

Answer:

4 moles of hydrogen (or the option corresponding to 4 moles of hydrogen, e.g., if the option is "4 moles of hydrogen", then that is the answer)