Answer:
2.54g
Step-by-step explanation:
Lead carbonate (PbCO3) can be decomposed to produce lead oxide (PbO) and carbon dioxide (CO2). The equation for this reaction is as follows:
PbCO3(s) -> PbO(s) + CO2(g)
Let's assume that the reaction yield is 95.7%, meaning that 95.7% of the theoretical amount of lead oxide produced in the reaction is actually obtained. To find the actual amount of lead oxide produced, we first need to find the theoretical amount of lead oxide produced by the reaction.
The theoretical amount of lead oxide can be calculated using the stoichiometry of the reaction, where the number of moles of reactants is balanced with the number of moles of products. If we assume that 2.50g of lead carbonate is decomposed, the number of moles of lead carbonate can be calculated as follows:
n = m/M
where n is the number of moles, m is the mass of the substance, and M is the molar mass of the substance. For lead carbonate, the molar mass is:
M = 207.19 g/mol
So, the number of moles of lead carbonate is:
n = 2.50 g / 207.19 g/mol = 0.01202 mol
Since the reaction is balanced, the number of moles of lead oxide produced should be equal to the number of moles of lead carbonate. The mass of lead oxide produced can be calculated using the number of moles and the molar mass of lead oxide:
m = n x M
where M = 223.20 g/mol is the molar mass of lead oxide. So, the mass of lead oxide produced is:
m = 0.01202 mol x 223.20 g/mol = 2.68 g
Since the reaction yield is 95.7%, the actual amount of lead oxide produced is:
actual_mass = 0.957 x 2.68 g = 2.54 g
So, approximately 2.54g of lead oxide will be produced by the decomposition of 2.50g of lead carbonate.