Question

Рассчитайте массу водорода, которая может получиться при растворении 112г железа в серной кислоте.

Answer 1

Дано:

m (Fe) – 112гр

Fe + H2SO4 

Найти:

m(H2) -?

 

Решение:

112гр                                     Хгр

   Fe + H2SO4 = FeSO4 + H2↑

m=56гр                               m=2гр

 

Составляем пропорцию:

112гр/56гр = Хгр/2гр

m(H2)= 112гр х 2гр /56гр = 4гр

Answer 2

Calculation of Hydrogen Mass from Iron Dissolution

To calculate the mass of hydrogen that can be produced when 112g of iron is dissolved in sulfuric acid, we need to consider the balanced chemical equation for the reaction between iron and sulfuric acid:

Fe + H2SO4 -> FeSO4 + H2

From the equation, we can see that 1 mole of iron reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas. To determine the mass of hydrogen gas produced, we need to convert the mass of iron to moles and then use the mole ratio from the balanced equation.

1. Calculate the moles of iron: - The molar mass of iron (Fe) is 55.845 g/mol. - Moles of iron = mass of iron / molar mass of iron - Moles of iron = 112g / 55.845 g/mol

2. Use the mole ratio to determine the moles of hydrogen gas: - From the balanced equation, the mole ratio of iron to hydrogen is 1:1. - Moles of hydrogen gas = Moles of iron

3. Calculate the mass of hydrogen gas: - The molar mass of hydrogen (H2) is 2.016 g/mol. - Mass of hydrogen gas = Moles of hydrogen gas * molar mass of hydrogen - Mass of hydrogen gas = Moles of iron * 2.016 g/mol

Now, let's calculate the mass of hydrogen gas that can be produced:

Moles of iron = 112g / 55.845 g/mol = 2.003 mol Moles of hydrogen gas = 2.003 mol Mass of hydrogen gas = 2.003 mol * 2.016 g/mol = 4.04 g

Therefore, when 112g of iron is dissolved in sulfuric acid, approximately 4.04 grams of hydrogen gas can be produced.

Please note that this calculation assumes ideal conditions and complete reaction conversion. In practice, the actual yield may be lower due to factors such as impurities, side reactions, and incomplete conversion.