Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide. PbCO3

Question

Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide.
PbCO3 (s) ——> PbO (s) + CO2 (g)
Calculate the mass (in grams) of each product formed when 2.50 grams of PbCO3 decomposes. Lead (II) carbonate decomposes to give lead (II) oxide and carbon dioxide.
PbCO3 (s) ——> PbO (s) + CO2 (g)
Calculate the mass (in grams) of each product formed when 2.50 grams of PbCO3 decomposes.
PbCO3 (s) ——> PbO (s) + CO2 (g)
Calculate the mass (in grams) of each product formed when 2.50 grams of PbCO3 decomposes.

Explanation / Answer

Numbre of moles of PbCO3 = 2.50 g / 267.21 g/mol = 0.00936 mol

From the balanced equation we can say that

1 mole of PbCO3 produces 1 mole of PbO so

0.00936 mole of PbCO3 will produce 0.00936 mole of PbO

mass of 1 mole of PbO = 223.2 g

so the msas of 0.00936 mole of PbO = 2.09 g

Therefore, the mass of PbO produced would be 2.09 g

From the balanced equation we can say that

1 mole of PbCO3 produces 1 mole of CO2 so

0.00936 mole of PbCO3 will produce

= 0.00936 mole of PbCO3 *( 1 mole of CO2 / 1 mole of PbCO3)

= 0.00936 mole of CO2

mass of 1 mole of CO2 = 44.01 g

so the mass of 0.00936 mole of CO2 = 0.412 g

Therefore, the mass of CO2 produced would be 0.412 g

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