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$\mathrm{1.45 \mathrm{~g}}$ of $\mathrm{P_4}$ was treated with $\mathrm{5.07 \mathrm{~g}}$ of $\mathrm{\mathrm{O}_2}$ . Calculate the mass of $\mathrm{\mathrm{P}_4 \mathrm{O}_{10}}$ formed in the given mixture:
Option: 1
1.45 g

Option: 2
5.07 g

Option: 3
3.32 g

Option: 4
10 g

Answers (1)

best_answer

$\mathrm{P_4(1.45 \mathrm{~g})=\frac{1.45}{31 \times 4} \text { moles of } P_4}$

$\mathrm{\text { 5.07g of } \mathrm{O}_2=\frac{5.07}{32} \text { moles of } \mathrm{O}_2}$

$\mathrm{\mathrm{P}_4+5 \mathrm{O}_2 \rightarrow \mathrm{P}_4 \mathrm{O}_{10}}$

From 5 moles of $\mathrm{Q_2}$ , moles of $\mathrm{P_4 q_0}$ obtain $\mathrm{=\frac{1}{5} \times \frac{5.07}{32}}$

$\mathrm{ =0.03169 \text { moles } }$
from 1 moles of $\mathrm{P_4}$ , moles of $\mathrm{P_4 q_0}$ obtained $\mathrm{=1}$ mol

from $\mathrm{\frac{1.45}{31 \times 4}}$ moles of $\mathrm{P_4, P_4 O_{10}}$ obtained $\mathrm{=\frac{1.45}{31 \times 4}}$

$\mathrm{ =0.0169 \text { moles } }$
$\mathrm{\therefore P_4}$ is the limiting reagent

Amount of $\mathrm{P_4 O_{10}}$ obtained $\mathrm{0.01169}$ moles

$\mathrm{ \begin{aligned} & =0.01169 \times 284 \\ & =3.32 \mathrm{~g} . \end{aligned} }$

Posted by

Ajit Kumar Dubey

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