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In the reaction $\mathrm{HNO}_3+\mathrm{P}_4 \mathrm{O}_{10} \rightarrow 4 \mathrm{HPO}_3+x$, the product $x$ is
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The correct answer is:
$\mathrm{N}_2 \mathrm{O}_5$
We know that, $4 \mathrm{HNO}_3+\mathrm{P}_4 \mathrm{O}_{10} \rightarrow 4 \mathrm{HPO}_3+2 \mathrm{~N}_2 \mathrm{O}_5$
The product is dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$
The product is dinitrogen pentaoxide $\left(\mathrm{N}_2 \mathrm{O}_5\right)$
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