Question

Необходимо приготовить сплав олова со свинцом, в котором массовая доля свинца 46,5% Какую массу оксида олова4 и оксида свинца2 надо взять чтобы при восстановлении их углем получился сплав массой 44,5г

Answer 1

1. Найдем массовую долю олова в сплаве:
w(Sn)=100%-46,5%=53,5%
2. Вычислим массу каждого металла в сплаве:
m(Pb)=89 г*0,465=41,385 г
m(Sn)=89 г*0,535=47,615 г
3. Составим уравнение реакции восстановления оксида свинца(2) углём:
PbO + C = Pb + CO
4. Вычислим количество вещества - свинца (Pb) по формуле n=m\M
n(Pb)=41,385г\ 207 г\моль≈0,2 моль
5. По уравнению реакции найдём количество вещества - оксида свинца(2) -PbO ( сравним коэффициенты : перед свинцом и оксидом свинца(2) стоит коэффициент 1 , так как 1=1, то n(Pb)=n(PbO):
По уравнению реакции:
n(PbO)=n(Pb)=0,2 моль
6. Вычислим массу оксида свинца(2) по формуле m=n*M:
m(PbO)=0,2 моль * 223 г\моль=44,6г
7. Составим уравнение восстановления углём оксида олова(4):
SnO₂ + C = Sn + CO₂
8. Найдём количество вещества олова(Sn) по формуле n=m\M:
n(Sn) = 47,615г \ 119 г\моль=0,4 моль
9. По уравнению реакции найдём количество вещества - оксида олова(4) - SnO₂ ( сравним коэффициенты : перед оловом и оксидом олова(4) коэффициент равен 1, так как 1=1 , то n(SnO₂)=n(Sn):
По уравнению реакции:
n(SnO₂)=n(Sn)=0,4 моль
10. Найдём массу оксида свинца(4)-SnO₂ по формуле m=n*M
m(SnO₂)=0,4 моль * 151 г\моль= 60,4г

Answer 2

Calculation of the Mass of Tin(IV) Oxide (SnO2) and Lead(II) Oxide (PbO) Required

To calculate the mass of tin(IV) oxide (SnO2) and lead(II) oxide (PbO) required to obtain an alloy with a mass of 44.5 grams and a lead mass fraction of 46.5%, we need to use the stoichiometry of the reaction and the molar masses of the compounds involved.

The balanced equation for the reaction between tin(IV) oxide and lead(II) oxide with carbon (coal) as the reducing agent is as follows:

SnO2 + 2PbO + C → Sn + 2Pb + CO2

From the balanced equation, we can see that the stoichiometric ratio between SnO2 and PbO is 1:2. This means that for every 1 mole of SnO2, we need 2 moles of PbO.

To calculate the mass of SnO2 and PbO required, we can follow these steps:

1. Calculate the molar mass of SnO2 and PbO: - The molar mass of SnO2 is the sum of the atomic masses of tin (Sn) and oxygen (O) in the compound. The atomic mass of Sn is 118.71 g/mol, and the atomic mass of O is 16.00 g/mol. Therefore, the molar mass of SnO2 is 118.71 g/mol + 2 * 16.00 g/mol = 150.71 g/mol. - The molar mass of PbO is the sum of the atomic masses of lead (Pb) and oxygen (O) in the compound. The atomic mass of Pb is 207.2 g/mol, and the atomic mass of O is 16.00 g/mol. Therefore, the molar mass of PbO is 207.2 g/mol + 16.00 g/mol = 223.2 g/mol.

2. Calculate the moles of PbO required: - We know that the mass fraction of lead (Pb) in the alloy is 46.5%. This means that for every 100 grams of the alloy, 46.5 grams will be lead. Therefore, for a 44.5-gram alloy, the mass of lead will be 44.5 g * (46.5 / 100) = 20.68 g. - To calculate the moles of PbO required, we divide the mass of lead by the molar mass of PbO: 20.68 g / 223.2 g/mol = 0.0927 mol.

3. Calculate the moles of SnO2 required: - Since the stoichiometric ratio between SnO2 and PbO is 1:2, we need twice the number of moles of PbO for SnO2. Therefore, the moles of SnO2 required will be 2 * 0.0927 mol = 0.1854 mol.

4. Calculate the mass of SnO2 and PbO required: - To calculate the mass of SnO2 required, we multiply the moles of SnO2 by its molar mass: 0.1854 mol * 150.71 g/mol = 27.89 g. - To calculate the mass of PbO required, we multiply the moles of PbO by its molar mass: 0.0927 mol * 223.2 g/mol = 20.68 g.

Therefore, to obtain an alloy with a mass of 44.5 grams and a lead mass fraction of 46.5%, you would need to take approximately 27.89 grams of tin(IV) oxide (SnO2) and 20.68 grams of lead(II) oxide (PbO) when they are reduced with carbon (coal).

Please note that these calculations are based on the given information and assumptions about the reaction conditions.