Question

Какая масса соли получается если в реакцию с гидроксидом железа(III) вступает 441 г серной кислоты . Выход продукта при этом составляет 98% от теоретически возможного? СРОЧНООООО!!!!!!

Answer 1

   Xg            141 g
2Fe(OH)3+3H2SO4--.Fe2(SO4)3+3H2O
 2*107        3*98
M(Fe(OH)3)=107 g/mol
M(H2SO4)=98g/mol
x/214= 141/294
X=102.6 g
m(пр)=102.6*98%/100%=100.5 g
ответ 100.5 г

Answer 2

Calculation of the Mass of Salt Produced

To calculate the mass of salt produced in the reaction between iron(III) hydroxide and sulfuric acid, we need to consider the stoichiometry of the reaction. The balanced chemical equation for the reaction is:

Fe(OH)3 + 3H2SO4 -> Fe2(SO4)3 + 3H2O

From the balanced equation, we can see that 1 mole of iron(III) hydroxide reacts with 3 moles of sulfuric acid to produce 1 mole of iron(III) sulfate and 3 moles of water.

To determine the mass of salt produced, we need to know the molar mass of iron(III) sulfate. The molar mass of Fe2(SO4)3 is calculated as follows:

Fe: 55.845 g/mol S: 32.06 g/mol O: 16.00 g/mol

Molar mass of Fe2(SO4)3 = (2 * 55.845) + (3 * 32.06) + (12 * 16.00) = 399.88 g/mol

Now, let's calculate the mass of salt produced using the given information.

Given: Mass of sulfuric acid (H2SO4) = 441 g Product yield = 98% of the theoretical yield

First, we need to calculate the theoretical yield of iron(III) sulfate using the molar ratio from the balanced equation.

Molar ratio of H2SO4 to Fe2(SO4)3 = 3:1

The molar mass of H2SO4 is calculated as follows:

H: 1.008 g/mol S: 32.06 g/mol O: 16.00 g/mol

Molar mass of H2SO4 = (2 * 1.008) + 32.06 + (4 * 16.00) = 98.09 g/mol

Now, we can calculate the moles of sulfuric acid used:

Moles of H2SO4 = Mass of H2SO4 / Molar mass of H2SO4 = 441 g / 98.09 g/mol = 4.494 mol

According to the stoichiometry of the balanced equation, 1 mole of Fe2(SO4)3 is produced for every 3 moles of H2SO4 used. Therefore, the moles of Fe2(SO4)3 produced can be calculated as follows:

Moles of Fe2(SO4)3 = (1/3) * Moles of H2SO4 = (1/3) * 4.494 mol = 1.498 mol

Finally, we can calculate the mass of iron(III) sulfate produced:

Mass of Fe2(SO4)3 = Moles of Fe2(SO4)3 * Molar mass of Fe2(SO4)3 = 1.498 mol * 399.88 g/mol = 599.14 g

However, the given information states that the product yield is 98% of the theoretical yield. Therefore, the actual mass of iron(III) sulfate produced is:

Actual mass of Fe2(SO4)3 = Product yield * Theoretical mass of Fe2(SO4)3 = 0.98 * 599.14 g = 587.32 g

Therefore, the mass of salt (iron(III) sulfate) produced in the reaction is approximately 587.32 grams.

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