# How many moles of water can be produced when "17.2 g B"_2"H"_6" undergoes complete combustion?

Mar 27, 2017

$1.24 \text{ mol}$ of ${H}_{2} O$

#### Explanation:

${B}_{2} {H}_{6} + {O}_{2} \to H B {O}_{2} + {H}_{2} O$

Balanced: ${B}_{2} {H}_{6} + 3 {O}_{2} \to 2 H B {O}_{2} + 2 {H}_{2} O$

Molar mass of ${B}_{2} {H}_{6} = 2 \left(10.81\right) + 6 \left(1.008\right) = 27.668 \text{ g/mol}$

Molar mass of ${H}_{2} O = 2 \left(1.008\right) + \left(15.999\right) = 18.015 \text{ g/mol}$

Moles of B_2H_6: (17.2g) /(27.668 " g/mol") = 0.6217 " mol"

From the balanced equation 0.6217 mol of ${B}_{2} {H}_{6}$ produces $2 \left(0.6217\right) = 1.24 \text{ mol}$ of ${H}_{2} O$

0.6217cancel("mol" B_2H_6)xx(2"mol" H_2O)/cancel(1"mol" B_2H_6)=1.24 "mol" H_2O (rounded to two significant figures)

May 19, 2018

The complete combustion of $\text{17.2 g B"_2"H"_6}$ produces $\text{2.01 mol H"_2"O}$.

#### Explanation:

Balanced equation

$\text{B"_2"H"_6("g") + "3O"_2("g")}$$\rightarrow$$\text{B"_2"O"_3("s") + "3H"_2"O(g)}$

We first need to determine mol $\text{B"_2"H"_6}$ by dividing its given mass by its molar mass $\left(\text{25.652 g/mol}\right)$. Divide by multiplying by the reciprocal of its molar mass (mol/g).

17.2color(red)cancel(color(black)("g B"_2"H"_6))xx(1"mol B"_2"H"_6)/(25.652color(red)cancel(color(black)("g B"_2"H"_6)))="0.671 mol B"_2"H"_6"

To determine mol $\text{H"_2"O}$ produced, multiply mol $\text{B"_2"H"_6}$ by the mol ratio between $\text{H"_2"O}$ and $\text{B"_2"H"_6}$ from the balanced equation, with $\text{H"_2"O}$ in the numerator.

0.671color(red)cancel(color(black)("mol B"_2"H"_6))xx(3"mol H"_2"O")/(1color(red)cancel(color(black)("mol B"_2"H"_6)))="2.01 mol H"_2"O"

The complete combustion of $\text{17.2 g B"_2"H"_6}$ produces $\text{2.01 mol H"_2"O}$.