In the reaction 2H_2 + O_2 to 2H_2O,if 6 g o | vedclass

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(C) The balanced chemical equation is: $2H_2 + O_2 	o 2H_2O$. Calculate the moles of reactants: Moles of $H_2 = \frac{6 \ g}{2 \ g/mol} = 3 \ mol$. M

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The balanced chemical equation is: $2H_2 + O_2 	o 2H_2O$. Calculate the moles of reactants: Moles of $H_2 = \frac{6 \ g}{2 \ g/mol} = 3 \ mol$. Moles of $O_2 = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$. According to the stoichiometry,$2 \ mol$ of $H_2$ reacts with $1 \ mol$ of $O_2$. Therefore,$3 \ mol$ of $H_2$ requires $\frac{1}{2} 	imes 3 = 1.5 \ mol$ of $O_2$. Since we have $2 \ mol$ of $O_2$,$O_2$ is the excess reagent. Remaining moles of $O_2 = 2 \ mol - 1.5 \ mol = 0.5 \ mol$. Mass of remaining $O_2 = 0.5 \ mol 	imes 32 \ g/mol = 16 \ g$.

In the reaction $2H_2 + O_2 \to 2H_2O$,if $6 \ g$ of $H_2$ combines with $64 \ g$ of $O_2$,find the mass of the excess reagent left in $g$.

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