Lead (II)carbonate decomposes to give lead (II)oxide and carbon dioxide: PbCO₃(s)→ PbO (s)+ CO₂(g) ________ grams of lead (II)oxide will be produced by the decomposition of 8.75 g of lead (II)carbonate?

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Quizplus · Accepted Answer

To solve this problem, we first need to balance the chemical equation and then use stoichiometry to find the mass of lead (II) oxide produced. The balanced equation for the decomposition of lead (II) carbonate (PbCO3) into lead (II) oxide (PbO) and carbon dioxide (CO2) is:$$ \text{PbCO}_3 (s) \rightarrow \text{PbO} (s) + \text{CO}_2 (g) $$Next, we calculate the molar mass of lead (II) carbonate and lead (II) oxide:- Molar mass of PbCO3 = 207.2 (Pb) + 12.01 (C) + 3*16.00 (O) = 267.21 g/mol- Molar mass of PbO = 207.2 (Pb) + 16.00 (O) = 223.2 g/molGiven 8.75 g of PbCO3, we find the moles of PbCO3:$$ \text{Moles of PbCO}_3 = \frac{8.75 \text{ g}}{267.21 \text{ g/mol}} = 0.0327 \text{ mol} $$Since the reaction is a 1:1 mole ratio between PbCO3 and PbO, the moles of PbO produced will also be 0.0327 mol. To find the mass of PbO produced:$$ \text{Mass of PbO} = 0.0327 \text{ mol} \times 223.2 \text{ g/mol} = 7.30 \text{ g} $$Therefore, 7.31 grams of lead (II) oxide will be produced by the decomposition of 8.75 g of lead (II) carbonate.

Lead (II)carbonate Decomposes to Give Lead (II)oxide and Carbon Dioxide