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

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(D) The balanced chemical equation is: $2H_2(g) + O_2(g) 	o 2H_2O(l)$.Calculate the moles of reactants:$n(H_2) = \frac{10 \ g}{2 \ g/mol} = 5 \ mol$.

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$4$

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(D) The balanced chemical equation is: $2H_2(g) + O_2(g) 	o 2H_2O(l)$.Calculate the moles of reactants:$n(H_2) = \frac{10 \ g}{2 \ g/mol} = 5 \ mol$.$n(O_2) = \frac{64 \ g}{32 \ g/mol} = 2 \ mol$.Identify the Limiting Reagent $(L.R.)$:According to the stoichiometry,$2 \ mol$ of $H_2$ requires $1 \ mol$ of $O_2$.So,$5 \ mol$ of $H_2$ would require $2.5 \ mol$ of $O_2$.Since we only have $2 \ mol$ of $O_2$,$O_2$ is the $L.R.$Calculate the product:Since $1 \ mol$ of $O_2$ produces $2 \ mol$ of $H_2O$,$2 \ mol$ of $O_2$ will produce $2 	imes 2 = 4 \ mol$ of $H_2O$.

$10 \ g$ 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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