Question

A scientist studying the reaction between decaborane and oxygen mixed 75.0 g of B10H18 with 145.0 g of O2. This reaction generates B2O3 and H2O as the only products. Compute how many grams of B2O3 are present after the reaction went to completion

Answer 1

Answer on Question #62272, Chemistry / General Chemistry

1. A scientist studying the reaction between decaborane and oxygen mixed $75.0\mathrm{g}$ of B10H18 with $145.0\mathrm{g}$ of O2. This reaction generates B2O3 and H2O as the only products. Compute how many grams of B2O3 are present after the reaction went to completion

Solution:

$$\mathrm{B}_{10}\mathrm{H}_{18} + 12\mathrm{O}_2 = 5\mathrm{B}_2\mathrm{O}_3 + 9\mathrm{H}_2\mathrm{O}$$ $$\begin{array}{l} \mathrm{M} \left(\mathrm{B}_{10}\mathrm{H}_{18}\right) = 10 \times 10.8 + 18 \times 1 = 126\ \mathrm{g/mol} \\ \mathrm{n} \left(\mathrm{B}_{10}\mathrm{H}_{18}\right) = \frac{75.0\ \mathrm{g}}{126\ \mathrm{g/mol}} = 0.595\ \mathrm{mol} \\ \mathrm{n} \left(\mathrm{O}_2\right) = \frac{145.0\ \mathrm{g}}{32\ \mathrm{g/mol}} = 4.531\ \mathrm{mol} \\ \end{array}$$

We need:

$\mathrm{n}(\mathrm{B}_{10}\mathrm{H}_{18}) : \mathrm{n}(\mathrm{O}_2) = 1:12$ (reaction),

we have: 1:7.6.

$\mathrm{B}_{10}\mathrm{H}_{18}$ - excess

$\mathrm{O}_2$ - deficit.

$$\begin{array}{l} \mathrm{M} \left(\mathrm{B}_2\mathrm{O}_3\right) = 70\ \mathrm{g/mol} \\ \mathrm{m} \left(\mathrm{B}_2\mathrm{O}_3\right) = \frac{145.0\ \mathrm{g} \times 5 \times 70\ \mathrm{g/mol}}{12 \times 32\ \mathrm{g/mol}} = 132.161\ \mathrm{g}. \\ \end{array}$$

Answer: $\mathrm{m}(\mathrm{B}_2\mathrm{O}_3) = 132.161\ \mathrm{g}$.

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