Question

How many moles of lead (II) chloride will be formed from a reaction between 6.5g PbO and 3.2g HCl ?
a.)0.011
b.)0.029
c.)0.044
d.)0.333

Answer 1

Hint: The number of moles formed will depend on PbO because it seems to be the limiting reagent in the reaction.
Number of moles = $\dfrac{{{\text{Weight of the substance(given)}}}}{{Molar{\text{ mass of the substance}}}}$

Step by step answer:
We can write the equation as-
$PbO + 2HCl \to PbC{l_2} + {H_2}O$
Thus, it is now clear that 1 mole of PbO will react with 2 moles of HCl to produce 1 mole of $PbC{l_2}$ and 1 mole of water. The PbO is the limiting reagent in the reaction.
The question comes in our mind that what a limiting reagent is now.
Limiting reagent is the reactant which is present in smaller quantities and the quantity of product is determined from it.
The PbO here being the limiting reagent because it is required in smaller quantities. So, the formation of $PbC{l_2}$ will be dependent on the moles of PbO.
From the question, we have reactant amount as-
Weight of PbO= 6.5g
Weight of HCl= 3.2g
Molar mass of PbO = 223g
Molar mass of HCl = 36.5g
Molar mass of $PbC{l_2}$ = 278g
We can say that 223g of PbO produces 278g of $PbC{l_2}$.
So, 6.5g of PbO will produce = $\dfrac{{278}}{{223}} \times 6.5 = 8.103g$
In terms of the number of moles, we can calculate as-
Number of moles of $PbC{l_2}$ = $\dfrac{{Weight{\text{ of PbC}}{{\text{l}}_2}}}{{Molar{\text{ mass of PbC}}{{\text{l}}_2}}}$
Number of moles of $PbC{l_2}$ = $\dfrac{{8.103g}}{{278g}} = 0.029$

So, option b.) is the correct answer.

Note:
We can find a number of moles of PbO and HCl.
We know, Number of moles of PbO = $\dfrac{{Weight{\text{ of PbO given}}}}{{Molar{\text{ mass of PbO}}}}$
Number of moles of PbO = $\dfrac{{6.5g}}{{223g}} = 0.0291$
Number of moles of HCl = $\dfrac{{Weight{\text{ of HCl}}}}{{Molar{\text{ mass of HCl}}}}$
Number of moles of HCl = $\dfrac{{3.2g}}{{36.5g}} = 0.087$
From the equation, we have seen that 1 mole of PbO produces 1 mole of $PbC{l_2}$ . So, in simple words, we can directly say that 0.029 moles of PbO will produce 0.029 moles of $PbC{l_2}$.