Answer :
First, we calculate the molar mass of diborane, [tex]$\mathrm{B_2H_6}$[/tex], using the atomic masses of boron and hydrogen. Given that the atomic mass of boron is approximately [tex]$10.81\,\mathrm{g/mol}$[/tex] and hydrogen is about [tex]$1.008\,\mathrm{g/mol}$[/tex], the molar mass is calculated as:
[tex]$$
\text{Molar mass of } \mathrm{B_2H_6} = 2(10.81\,\mathrm{g/mol}) + 6(1.008\,\mathrm{g/mol}) = 21.62\,\mathrm{g/mol} + 6.048\,\mathrm{g/mol} = 27.668\,\mathrm{g/mol}
$$[/tex]
Next, the number of moles of [tex]$\mathrm{B_2H_6}$[/tex] in [tex]$32.1\,\mathrm{g}$[/tex] is determined by:
[tex]$$
\text{Moles of } \mathrm{B_2H_6} = \frac{32.1\,\mathrm{g}}{27.668\,\mathrm{g/mol}} \approx 1.16\,\mathrm{mol}
$$[/tex]
The balanced chemical reaction is:
[tex]$$
\mathrm{B_2H_6} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{HBO_2} + 2\,\mathrm{H_2O}
$$[/tex]
This equation shows that [tex]$1$[/tex] mole of [tex]$\mathrm{B_2H_6}$[/tex] reacts with [tex]$3$[/tex] moles of [tex]$\mathrm{O_2}$[/tex]. Therefore, the moles of [tex]$\mathrm{O_2}$[/tex] required are:
[tex]$$
\text{Moles of } \mathrm{O_2} = 1.16\,\mathrm{mol} \times 3 \approx 3.48\,\mathrm{mol}
$$[/tex]
Since the molar mass of oxygen gas, [tex]$\mathrm{O_2}$[/tex], is [tex]$32.00\,\mathrm{g/mol}$[/tex], the mass of [tex]$\mathrm{O_2}$[/tex] needed is:
[tex]$$
\text{Mass of } \mathrm{O_2} = 3.48\,\mathrm{mol} \times 32.00\,\mathrm{g/mol} \approx 111.38\,\mathrm{g}
$$[/tex]
Rounding to the appropriate significant figures, we conclude that approximately [tex]$111\,\mathrm{g}$[/tex] of oxygen is required.
[tex]$$
\text{Molar mass of } \mathrm{B_2H_6} = 2(10.81\,\mathrm{g/mol}) + 6(1.008\,\mathrm{g/mol}) = 21.62\,\mathrm{g/mol} + 6.048\,\mathrm{g/mol} = 27.668\,\mathrm{g/mol}
$$[/tex]
Next, the number of moles of [tex]$\mathrm{B_2H_6}$[/tex] in [tex]$32.1\,\mathrm{g}$[/tex] is determined by:
[tex]$$
\text{Moles of } \mathrm{B_2H_6} = \frac{32.1\,\mathrm{g}}{27.668\,\mathrm{g/mol}} \approx 1.16\,\mathrm{mol}
$$[/tex]
The balanced chemical reaction is:
[tex]$$
\mathrm{B_2H_6} + 3\,\mathrm{O_2} \rightarrow 2\,\mathrm{HBO_2} + 2\,\mathrm{H_2O}
$$[/tex]
This equation shows that [tex]$1$[/tex] mole of [tex]$\mathrm{B_2H_6}$[/tex] reacts with [tex]$3$[/tex] moles of [tex]$\mathrm{O_2}$[/tex]. Therefore, the moles of [tex]$\mathrm{O_2}$[/tex] required are:
[tex]$$
\text{Moles of } \mathrm{O_2} = 1.16\,\mathrm{mol} \times 3 \approx 3.48\,\mathrm{mol}
$$[/tex]
Since the molar mass of oxygen gas, [tex]$\mathrm{O_2}$[/tex], is [tex]$32.00\,\mathrm{g/mol}$[/tex], the mass of [tex]$\mathrm{O_2}$[/tex] needed is:
[tex]$$
\text{Mass of } \mathrm{O_2} = 3.48\,\mathrm{mol} \times 32.00\,\mathrm{g/mol} \approx 111.38\,\mathrm{g}
$$[/tex]
Rounding to the appropriate significant figures, we conclude that approximately [tex]$111\,\mathrm{g}$[/tex] of oxygen is required.
