K + H2O --> KOH + H2

1. How many grams of KOH are produced if 224 g of H2O are used?
2. How many moles of K are used if 20 moles of H2O react?
3. How many grams of H2O are used if 3678 grams of H2 are produced?

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

they are both synthetic elements

explanation:

d

the product of the pressure and volume is also constant