According to the following reaction, how many grams of potassium sulfate will be formed upon the complete reaction of 23.8 grams of potassium hydroxide with excess potassium hydrogen sulfate? potassium hydrogen sulfate (aq) + potassium hydroxide (aq) potassium sulfate (aq) + water (l)

The mass of potassium sulfate that can be produced is 73.88 grams

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of KOH = 23.8 g

Molar mass of KOH = 56.1 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of KOH and potassium hydrogen sulfate follows:

As, potassium hydrogen sulfate is present in excess. It is considered as an excess reagent.

KOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of KOH produces 1 mole of potassium sulfate

So, 0.424 moles of KOH will produce = of potassium sulfate

Now, calculating the mass of potassium sulfate from equation 1, we get:

Molar mass of potassium sulfate = 174.26 g/mol

Moles of potassium sulfate = 0.424 moles

Putting values in equation 1, we get:

Hence, the mass of potassium sulfate that can be produced is 73.88 grams

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n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.