Answer:
1. 696.9g of KOH
2. 20moles of K
3. 66204g of H2O
Explanation:
First, we need to generate a balanced equation for the reaction. This is illustrated below:
2K + 2H2O —> 2KOH + H2
Next, we'll find the molar mass and mass of the reactants and products as shown below:
Molar Mass of K = 39g/mol
Mass of K from the balanced equation = 2x39 = 78g
Molar Mass of H2O = (2x1) + 16 = 18g/mol
Mass of H2O from the balanced equation = 18 x 2 = 36g
Molar Mass of KOH = 39 + 16 + 1 = 56g/mol
Mass of KOH from the balanced equation = 2 x 56 = 112g
Molar Mass of H2 = 2x1 = 2g/mol
1. 2K + 2H2O —> 2KOH + H2
From the balanced equation,
36g of H20 produced 112g of KOH.
Therefore, 224g of H2O will produce = (224 x 112)/36 = 696.9g of KOH
2. 2K + 2H2O —> 2KOH + H2
From the balanced equation,
2moles of K required 2moles of H2O.
Therefore, Xmol of K will require 20moles of H2O i.e
Xmol of K = (20 x 2) / 2 = 20moles
3. 2K + 2H2O —> 2KOH + H2
From the balanced equation,
36g of H2O produced 2g of H2.
Therefore,
Xg of H2O will produce 3678g of H2 i.e
Xg of H2O = (3678 x 36)/2 = 66204g
