Step 1 - Finding the relation in moles in the reaction equation
As we have previously seen working on a similar problem, the bigger numbers that come before the formula of the substance expresses a proportion in moles between all the reactants and the products.
The reaction given in this exercise is:
[tex]2K_{(s)}+2H_2O_{(l)}\to H_{2(g)}+2\text{KOH}_{(aq)}[/tex]Note that 2 moles of K react with 2 moles of H2O, producing 1 mole of H2 and 2 moles of KOH.
Step 2 - Converting a relation in moles to a relation in grams
We can convert this relation in moles to a relation in grams by multiplying each number of moles by the molar mass of each substance. The molar masses are:
[tex]\begin{gathered} K\to39\text{ g/mol} \\ H_2O\to18\text{ g/mol} \\ H_2\to2\text{ g/mol} \\ \text{KOH}\to56\text{ g/mol} \end{gathered}[/tex]Converting to grams:
[tex]\begin{gathered} K\to2\times39=78g \\ H_2O\to2\times18=36g \\ H_2\to1\times2=2g \\ \text{KOH}\to2\times56=112g \end{gathered}[/tex]We can state thus that 78g of K react with 36g of water, producing 2g of H2 and 112g of KOH. This is a fixed proportion, like a cake recipe.
Step 3 - Using the information to answer the exercise
1) Note that 2 moles of K produces 2 moles of KOH. Therefore, we would need exactly 2 moles of K.
2) Note that 2 moles of K, as we have previously calculated in step 2, corresponds to 78 g.
3) Note that 2 moles of K produces 1 mole of H2. Since this is a fixed proportion, we can set the following relation:
[tex]\begin{gathered} 2\text{ moles of K produces --- 1 mole of H2} \\ 3\text{ moles of K would produce -- x} \\ \\ x=\frac{3\times1}{2}\text{=1}.5\text{ moles of H2} \end{gathered}[/tex]It would be produced thus 1.5 moles of H2.
4) As we have seen in 3), we would produce 1.5 moles of H2. Since its molar mass is 2g/mol, we would produce, in grams:
[tex]m=1.5\times2=3g[/tex]3g of H2 would be produced.
5) Note that 2 moles of H2O produce 1 mole of H2. In grams, this corresponds to 36g of water produce 2g of H2.
