Ν(h2so4)=0.2 моль m(koh)=28 г m(k2so4)-?

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вопрос

To find the mass of potassium sulfate (K2SO4) produced from the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH), we first need to determine the limiting reagent.

First, we need to write and balance the chemical equation for the reaction between H2SO4 and KOH:

H2SO4 + 2 KOH -> K2SO4 + 2 H2O

Now we need to calculate the number of moles of KOH used in the reaction:

n(KOH) = m(KOH) / M(KOH)
n(KOH) = 28g / 56.11g/mol
n(KOH) ≈ 0.5 mol

From the balanced chemical equation, we can see that 1 mole of K2SO4 is produced for every 2 moles of KOH used. Therefore, the number of moles of K2SO4 produced will be half of the moles of KOH used:

n(K2SO4) = n(KOH) / 2
n(K2SO4) = 0.5 mol / 2
n(K2SO4) = 0.25 mol

Finally, we can calculate the mass of K2SO4 produced:

m(K2SO4) = n(K2SO4) M(K2SO4)
m(K2SO4) = 0.25 mol 174.26 g/mol
m(K2SO4) = 43.57 g

Therefore, the mass of potassium sulfate produced from the reaction is approximately 43.57 grams.