A mixture containing 64.0 g of H_2 and 64.0 | vedclass

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(D) $n_{H_2} = \frac{64}{2} = 32 \ mol$ $n_{O_2} = \frac{64}{32} = 2 \ mol$ According to the reaction $2H_2 + O_2 \rightarrow 2H_2O$,$1 \ mol$ of $O_2

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Both $B$ and $C$

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(D) $n_{H_2} = \frac{64}{2} = 32 \ mol$ $n_{O_2} = \frac{64}{32} = 2 \ mol$ According to the reaction $2H_2 + O_2 ightarrow 2H_2O$,$1 \ mol$ of $O_2$ requires $2 \ mol$ of $H_2$. Therefore,$2 \ mol$ of $O_2$ requires $4 \ mol$ of $H_2$. Since $H_2$ is present in excess $(32 \ mol > 4 \ mol)$,$O_2$ is the limiting reagent. $A$. Incorrect. $B$. Correct,$O_2$ is the limiting reagent. $C$. $2 \ mol$ of $O_2$ produces $4 \ mol$ of $H_2O = 4 	imes 18 = 72 \ g \ H_2O$. Moles of $H_2$ left $= 32 - 4 = 28 \ mol$. Mass of $H_2$ left $= 28 \ mol 	imes 2 \ g/mol = 56 \ g$. Thus,the mixture contains $72 \ g \ H_2O$ and $56 \ g \ H_2$. Correct.

$A$ mixture containing $64.0 \ g$ of $H_2$ and $64.0 \ g$ of $O_2$ is reacted so that water is formed as follows:
$2H_2 + O_2 \rightarrow 2H_2O$
Which of the following statements is/are correct?

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$A$ mixture containing $64.0 \ g$ of $H_2$ and $64.0 \ g$ of $O_2$ is reacted so that water is formed as follows: $2H_2 + O_2 \rightarrow 2H_2O$ Which of the following statements is/are correct?