56 g of CaO has been mixed with 63 g of HNO | vedclass

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(D) The balanced chemical equation is:$CaO + 2HNO_3 \longrightarrow Ca(NO_3)_2 + H_2O$Molar masses:$CaO = 40 + 16 = 56 \ g/mol$$HNO_3 = 1 + 14 + 48 =

Vedclass · Accepted Answer

$82$

Vedclass · Answer

(D) The balanced chemical equation is:$CaO + 2HNO_3 \longrightarrow Ca(NO_3)_2 + H_2O$Molar masses:$CaO = 40 + 16 = 56 \ g/mol$$HNO_3 = 1 + 14 + 48 = 63 \ g/mol$$Ca(NO_3)_2 = 40 + 2(14 + 48) = 40 + 124 = 164 \ g/mol$Given amounts:$CaO = 56 \ g = 1 \ mol$$HNO_3 = 63 \ g = 1 \ mol$According to the stoichiometry, $1 \ mol$ of $CaO$ requires $2 \ mol$ of $HNO_3$. Since we only have $1 \ mol$ of $HNO_3$, $HNO_3$ is the limiting reagent.From the equation, $2 \ mol$ of $HNO_3$ produces $1 \ mol$ of $Ca(NO_3)_2$ $(164 \ g)$.Therefore, $1 \ mol$ of $HNO_3$ will produce:$\frac{164 \ g}{2} = 82 \ g$ of $Ca(NO_3)_2$.

$56 \ g$ of $CaO$ has been mixed with $63 \ g$ of $HNO_3$. The amount of $Ca(NO_3)_2$ formed is: (in $g$)

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