Question

M(Ka2SO4)=8,7г. n(KOH)-?

Answer 1

2KOH+H2SO4->K2SO4+H2O
n=0,1 моль

Answer 2

To find the amount of KOH (potassium hydroxide), we can use stoichiometry and the molar mass of potassium sulfate (K2SO4). The molar mass of K2SO4 is the sum of the molar masses of its constituent elements:

Molar mass of K2SO4 = 2 * molar mass of K + molar mass of S + 4 * molar mass of O = 2 * 39.10 g/mol + 32.07 g/mol + 4 * 16.00 g/mol = 174.26 g/mol

Given that the mass of K2SO4 (M) is 8.7 g, we can use the molar mass to find the number of moles of K2SO4:

Number of moles = Mass / Molar mass = 8.7 g / 174.26 g/mol ≈ 0.0499 mol

Now, we'll use the balanced chemical equation for the reaction between K2SO4 and KOH:

K2SO4 + 2 KOH → 2 K2O + H2SO4

From the equation, we can see that 1 mole of K2SO4 reacts with 2 moles of KOH. Therefore, the number of moles of KOH needed is twice the number of moles of K2SO4:

Number of moles of KOH = 2 * Number of moles of K2SO4 ≈ 2 * 0.0499 mol ≈ 0.0998 mol

Finally, let's find the mass of KOH using its molar mass:

Molar mass of KOH = 39.10 g/mol + 15.999 g/mol + 1.007 g/mol ≈ 56.11 g/mol

Mass of KOH = Number of moles of KOH * Molar mass of KOH ≈ 0.0998 mol * 56.11 g/mol ≈ 5.60 g

Therefore, the mass of KOH required is approximately 5.60 grams.