10 g of hydrogen and 64 g of oxygen were fi | vedclass

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(B) The balanced chemical equation for the reaction is: $2 H_2(g) + O_2(g) \rightarrow 2 H_2 O(l)$.Calculate the number of moles for the reactants:$n(

Vedclass · Accepted Answer

$4$

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(B) The balanced chemical equation for the reaction is: $2 H_2(g) + O_2(g) ightarrow 2 H_2 O(l)$.Calculate the number of moles for the reactants:$n(H_2) = \frac{10 \ g}{2 \ g/mol} = 5 \ mol$.$n(O_2) = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$.According to the stoichiometry,$1 \ mol$ of $O_2$ reacts with $2 \ mol$ of $H_2$ to produce $2 \ mol$ of $H_2 O$.Here,$O_2$ is the limiting reagent because it will be consumed completely first ($2 \ mol$ of $O_2$ requires $4 \ mol$ of $H_2$,and we have $5 \ mol$ of $H_2$).Since $1 \ mol$ of $O_2$ produces $2 \ mol$ of $H_2 O$,then $2 \ mol$ of $O_2$ will produce $2 	imes 2 = 4 \ mol$ of $H_2 O$.

$10 \ g$ of hydrogen and $64 \ g$ of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be ........... $mol$.

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