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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 ?
A
0.011
B
0.029
C
0.044
D
0.333
Detailed Solution
$PbO + 2HCl \rightarrow PbCl_2 + H_2O$
1 mole of PbO reacts with 2 moles of HCl to give 1 mole of $PbCl_2$.
Molar mass of PbO $= 207 + 16 = 223$ g mol$^{-1}$ (taken as about 224 in the source).
Moles of PbO $= \dfrac{6.5}{224} = 0.029$ mole
Moles of HCl $= \dfrac{3.2}{36.5} = 0.087$ mole
HCl required for 0.029 mole of PbO $= 2 \times 0.029 = 0.058$ mole, which is less than the 0.087 mole available.
So HCl is in excess and PbO is the limiting reagent.
Moles of $PbCl_2$ formed = moles of PbO = 0.029
Thus, 0.029 mole of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl.
1 mole of PbO reacts with 2 moles of HCl to give 1 mole of $PbCl_2$.
Molar mass of PbO $= 207 + 16 = 223$ g mol$^{-1}$ (taken as about 224 in the source).
Moles of PbO $= \dfrac{6.5}{224} = 0.029$ mole
Moles of HCl $= \dfrac{3.2}{36.5} = 0.087$ mole
HCl required for 0.029 mole of PbO $= 2 \times 0.029 = 0.058$ mole, which is less than the 0.087 mole available.
So HCl is in excess and PbO is the limiting reagent.
Moles of $PbCl_2$ formed = moles of PbO = 0.029
Thus, 0.029 mole of lead (II) chloride will be formed from a reaction between 6.5 g of PbO and 3.2 g of HCl.