Question

Consider the reaction of tin with potassium hydroxide and water. Sn (s) 2KOH (aq) 4H2O (l) K2Sn(OH)6 (s) 2H2 (g) Determine the limiting reactant in a mixture containing 122 g of Sn, 134 g of KOH, and 63.9 g of H2O. Calculate the maximum mass (in grams) of potassium hydroxystannate, K2Sn(OH)6, that can be produced in the reaction. The limiting reactant is:

Answer 1

The limiting reactant is: H₂O

265.3 g of K₂Sn(OH)₆ are formed in the reaction.

Step-by-step explanation:

The reaction is:

Sn(s) + 2KOH (aq) + 4H₂O(l) → K₂Sn(OH)₆ (s) + 2H₂ (g)

The first step to solve is to determine the moles of each reactant:

We controlled that reaction is ballanced.

122 g . 1mol / 118.71g = 1.03 moles of Sn

134g . 1mol /56.1g = 2.39 moles of KOH

63.9 g .1mol /18g = 3.55 mol of water

Stoichiometry is 1:2:4.

Sn is the lowest reactant and water, the highest, but I can see, that water is the limiting.

1 mol of Sn needs 4 moles of H₂O to react

Then, 1.03 moles of Sn may react to (1.03 . 4)/1 = 4.12 moles.

We only have 3.55 moles. It's ok.

2 moles of KOH need 4 moles of H₂O to react

Then, 2.39 moles of KOH may react to (2.39 . 4) /2 = 4.78 moles.

We only have 3.55, there's no enough water.

So 4 moles of water can produce 1 mol of potassium hydroxystannate

Then, 3.55 moles of H₂O may produce (3.55 . 1)/4 = 0.8875 moles.

We convert moles to mass: 0.08875 mol . 298.91g /1mol =265.3g