Al + 3 NaOH → Al(OH)3 + 3 Na Using the above balanced chemical equation, how many grams of Al are needed to react with 4.6 grams of NaOH?

Answer 1

For this question we will be using a periodic table for the molecular weights of Al and NaOH. We will also be using stoichometry for the ratios of the compounds based on the balanced equation.

Molecular weight of Al: 26.98 g/mol

Molecular weight of NaOH: 40.00 g/mol

Ratio of Al fo NaOH: 1 Al is needed for every 3 NaOH

We start with the given information (grams of NaOH) and then use the ratio (1 mol Al:3 mol NaOH) and molecular weights to calculate the desired information, which is the grams of Al.

4.6 g NaOH × (1 mol NaOH/40.00 g NaOH) × (1 mol Al/3 mol NaOH) × (26.98 g Al/1 mol Al) =

We enter this in the calculator as numerators divided by denominators or (4.6 × 26.98)/(40.00 × 3) =

And our answer is 1.03 which we round to 2 significant figures (in the given) to get our answer: 1.0 g Al.