0.1 mol of XeF_6 is treated with 1.8 g of w | vedclass

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(A) The molar mass of $H_2O$ is $18 \ g/mol$. Given mass of $H_2O = 1.8 \ g$. Number of moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$. The

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$XeOF_4$

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(A) The molar mass of $H_2O$ is $18 \ g/mol$. Given mass of $H_2O = 1.8 \ g$. Number of moles of $H_2O = \frac{1.8 \ g}{18 \ g/mol} = 0.1 \ mol$. The balanced chemical equation for the reaction of $XeF_6$ with $H_2O$ in a $1:1$ molar ratio is: $XeF_6 + H_2O \rightarrow XeOF_4 + 2HF$. Since $0.1 \ mol$ of $XeF_6$ reacts with $0.1 \ mol$ of $H_2O$,the product formed is $XeOF_4$.

$0.1 \ mol$ of $XeF_6$ is treated with $1.8 \ g$ of water. The product obtained is

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