Answer:
The limiting reactant is: H₂O
265.3 g of K₂Sn(OH)₆ are formed in the reaction.
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
