Question

For the following reaction, 42.2 grams of potassium hydrogen sulfate are allowed to react with 21.4 grams of potassium hydroxide. potassium hydrogen sulfate (aq) potassium hydroxide (aq) potassium sulfate (aq) water (l) What is the maximum amount of potassium sulfate that can be formed

Answer 1

The maximum amount of potassium sulfate formed is dictated by the limiting reactant, which in this case is KOH. Approximately 66.4 grams of potassium sulfate can be produced from the reaction of 21.4 grams of KOH with 42.2 grams of potassium hydrogen sulfate.

Step-by-step explanation:

To determine the maximum amount of potassium sulfate that can be formed from the reaction of potassium hydrogen sulfate and potassium hydroxide, we use stoichiometry based on the balanced chemical equation. Since the balanced equation is not provided, we can assume it is:

KHSO4(aq) + KOH(aq) → K2SO4(aq) + H2O(l)

Finding the mole ratios from the molecular weights, we'd say that 21.4 grams of KOH is less than the stoichiometric amount needed to fully react with 42.2 grams of potassium hydrogen sulfate. Therefore, KOH is the limiting reactant. The maximum amount of potassium sulfate is dictated by the amount of KOH that can react:

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• 1 mole of KOH has a mass of approximately 56.11 grams.
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• 21.4 grams of KOH is thus 0.381 moles of KOH.
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• According to the balanced equation, 1 mole of KOH reacts to produce 1 mole of K2SO4.
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• Therefore, a maximum of 0.381 moles of K2SO4 can be produced.
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• The molar mass of K2SO4 is approximately 174.26 grams/mole.
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• The maximum mass of K2SO4 that can be produced is 0.381 moles times 174.26 grams/mole, which equals approximately 66.4 grams.

Answer 2

53.99g

Step-by-step explanation:

Step 1:

The balanced equation for the reaction. This is given below:

KHSO4(aq) + KOH(aq) —> K2SO4(aq) + H2O(l)

Step 2:

Determination of the masses of KHSO4 and KOH that reacted and the mass of K2SO4 produced from the balanced equation.

This is illustrated below:

Molar mass of KHSO4 = 39 + 1 + 32 + (16x4) = 136g/mol

Mass of KHSO4 from the balanced equation = 1 x 136 = 136g

Molar mass of KOH = 39 + 16 + 1 = 56g/mol

Mass of KOH from the balanced equation = 1 x 56 = 56g

Molar mass of K2SO4 = (39x2) + 32 + (16x4) = 174g/mol

Mass of K2SO4 from the balanced equation = 1 x 174 = 174g.

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH to produce 174g of K2SO4

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above, 136g of KHSO4 reacted with 56g of KOH.

Therefore, 42.2g of KHSO4 will react with = (42.2 x 56)/136 = 17.38g of KOH.

From the above calculations, we can see that only 17.38g out of 21.4g of KOH given was needed to react completely with 42.2g of KHSO4.

Therefore, KHSO4 is the limiting reactant and KOH is the excess reactant.

Step 4:

Determination of the maximum mass of K2SO4 produced from the reaction.

In this case, the limiting reactant will be used as all of it is used up in the reaction. The limiting reactant is KHSO4 and the maximum amount of K2SO4 produced can be obtained as follow:

From the balanced equation above, 136g of KHSO4 reacted to produce 174g of K2SO4.

Therefore, 42.2g of KHSO4 will react to produce = (42.2 x 174)/136 = 53.99g of K2SO4.

Therefore, the maximum amount of K2SO4 produced is 53.99g.