Question

Вычислите массу водорода - газа полученного при взаимодействии 10,8 г алюминия с раствором серной кислоты

Answer 1

Дано:

m(Al)=10,8 г

Найти:

m(H2)-?

Решение:

2Al+3H2SO4=Al2(SO4)3+3H2

n(Al)=m/M =10,8 г: 27г/моль = 0,4 моль

M(Al)= 27 г/моль

n(Al)=n(H2)= 0,6 моль

m(H2)= n*M = 0,6 моль *2 г/моль =1,2 г

M(H2)= 1*2=2 г/моль

Ответ: m(H2)- 1,2 г

Answer 2

Calculating the Mass of Hydrogen Gas Produced

To calculate the mass of hydrogen gas produced when 10.8 g of aluminum reacts with sulfuric acid, we can use the balanced chemical equation for the reaction between aluminum and sulfuric acid:

2 Al(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)

From the balanced chemical equation, we can see that 2 moles of aluminum react with 3 moles of sulfuric acid to produce 3 moles of hydrogen gas. Now, let's calculate the amount of hydrogen gas produced.

First, we need to find the number of moles of aluminum: Molar mass of aluminum (Al) = 27 g/mol Number of moles of aluminum = Mass of aluminum / Molar mass of aluminum Number of moles of aluminum = 10.8 g / 27 g/mol Number of moles of aluminum = 0.4 moles

According to the balanced chemical equation, 2 moles of aluminum produce 3 moles of hydrogen gas. Therefore, 0.4 moles of aluminum will produce: Number of moles of hydrogen gas = (3 moles / 2 moles) * 0.4 moles Number of moles of hydrogen gas = 0.6 moles

Now, let's calculate the mass of hydrogen gas produced: Molar mass of hydrogen (H2) = 2 g/mol Mass of hydrogen gas = Number of moles of hydrogen gas * Molar mass of hydrogen Mass of hydrogen gas = 0.6 moles * 2 g/mol Mass of hydrogen gas = 1.2 g

Therefore, the mass of hydrogen gas produced when 10.8 g of aluminum reacts with sulfuric acid is 1.2 g.

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