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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?
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?
Updated On: May 5, 2024
- 0.339
- 0.011
- 0.029
- 0.044
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The Correct Option is C
Solution and Explanation
The reaction equation is
$PbO + 2HCl → PbCl_2 + H_2O$.
Molar mass of $PbO =223\,g\,mol^{-1}$.
Now, moles of $PbO=\frac{6.5}{223}=0.029\,mol$
and moles of $HCl=\frac{3.2}{36.5}=0.088\,mol$
$1\, mol$ of $PbO$ reacts with 2 moles of $HCl$.
Therefore, $PbO$ is a limiting reagent.
$0.029 \,mol$ of $PbO$ will produce $PbCl_2 = 0.029\, mol$
$PbO + 2HCl → PbCl_2 + H_2O$.
Molar mass of $PbO =223\,g\,mol^{-1}$.
Now, moles of $PbO=\frac{6.5}{223}=0.029\,mol$
and moles of $HCl=\frac{3.2}{36.5}=0.088\,mol$
$1\, mol$ of $PbO$ reacts with 2 moles of $HCl$.
Therefore, $PbO$ is a limiting reagent.
$0.029 \,mol$ of $PbO$ will produce $PbCl_2 = 0.029\, mol$
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