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Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?

When the equation is PbCO3(s) + 2H(+)(aq) = Pb(2+)(aq) + H2O(l) + CO2(g)

Please explain in detail, step by step

Verified Answer :

MissPhiladelphia

Calculating for the moles of H+
1.0 L x (1.00 mole / 1 L ) = 1 mole H+

From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3

Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
0.5 moles PbCO3 x (267 g PbCO3 / 1 mole PbCO3) = 84.5 g PbCO3

Answer: 6 grams of lead carbonate wlll get easily dissolve in 1 L of 1.0 M \(H^+\) solution.

Explanation:

\(PbCO_3(s)+2H^+(aq)\rightarrow Pb^2+(aq)+H_2O(l)+CO_2(g)\)

Moles of \(H^+\) (n):

\(1.0 M=\frac{\text{mole of }H^+}{\text{Volume of the solution is L}}=\frac{n}{1 L}\)

n = 1 mole

According to reaction 2 moles of \(H^+\) dissolves 1 mole of \(PbCO_3\) then , 1 mole will dissolve ;\(\frac{1}{2}\times 1\) moles of \(PbCO_3\) that is 0.5 moles

Mass of the compund:

=Number of moles of compound × Molar mass of compound

Mass of dissolved \(PbCO_3\):

=\(0.5 mole\times 267.21 g/mol=133.605 g\)

1 L of 1.0 M \(H^+\) solution can dissolve 133.605 grams of lead carbonate. So, 6 grams of lead carbonate wlll get easily dissolve in 1 L of 1.0 M \(H^+\) solution.

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