How many moles of lead (II) chloride will be | vedclass

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Question

(B) The balanced chemical equation is: $PbO + 2HCl \rightarrow PbCl_2 + H_2O$Calculate the molar masses:$PbO = 207.2 + 16.0 = 223.2 \ g/mol$$HCl = 1.0

Vedclass · Accepted Answer

$0.029$

Vedclass · Answer

(B) The balanced chemical equation is: $PbO + 2HCl ightarrow PbCl_2 + H_2O$Calculate the molar masses:$PbO = 207.2 + 16.0 = 223.2 \ g/mol$$HCl = 1.0 + 35.5 = 36.5 \ g/mol$Calculate the moles of reactants:$Moles \ of \ PbO = \frac{6.5 \ g}{223.2 \ g/mol} \approx 0.0291 \ mol$$Moles \ of \ HCl = \frac{3.2 \ g}{36.5 \ g/mol} \approx 0.0877 \ mol$According to the stoichiometry,$1 \ mol$ of $PbO$ requires $2 \ mol$ of $HCl$. For $0.0291 \ mol$ of $PbO$,we need $0.0291 	imes 2 = 0.0582 \ mol$ of $HCl$.Since we have $0.0877 \ mol$ of $HCl$,$PbO$ is the limiting reagent.The moles of $PbCl_2$ formed will be equal to the moles of $PbO$ consumed,which is $0.0291 \ mol$.

How many moles of lead $(II)$ chloride will be formed from a reaction between $6.5 \ g$ of $PbO$ and $3.2 \ g$ of $HCl$?

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