To determine the mole ratio between water (H₂O) and sulfuric acid (H₂SO₄) in the balanced chemical equation:
[tex]\[
3 H_2SO_4 + 2 B(OH)_3 \rightarrow B_2(SO_4)_3 + 6 H_2O
\][/tex]
we follow these steps:
1. Identify the coefficients of the substances in question from the balanced equation. Here, the coefficient of sulfuric acid (H₂SO₄) is 3, and the coefficient of water (H₂O) is 6.
2. The mole ratio is determined by the ratio of these coefficients. We find the ratio by dividing the coefficient of H₂O by the coefficient of H₂SO₄:
[tex]\[
\text{Mole ratio} = \frac{\text{Coefficient of } H_2O}{\text{Coefficient of } H_2SO_4} = \frac{6}{3}
\][/tex]
3. Simplify the division:
[tex]\[
\frac{6}{3} = 2
\][/tex]
4. Hence, the mole ratio between water (H₂O) and sulfuric acid (H₂SO₄) is:
[tex]\[
\frac{2 \text{ mol H}_2O}{1 \text{ mol H}_2SO_4}
\][/tex]
This means that for every 1 mole of sulfuric acid, 2 moles of water are produced.
