Question

A student reacts 0.506 grams of aluminum with excess KOH and H2O according to reaction one from the lab manual. Enter your answers in numerical format

How many moles of aluminum was reacted?

_______ moles Al

How many moles of K[Al(OH)4] will theoretically be produced?

_______ moles K[Al(OH)4]

Answer 1

Sol:-

Balanced Chemical Equation for the formation of K[Al(OH)₄] is :

2 Al + 2 KOH + 6 H₂O -----------> 2 K[Al(OH)₄] + 3 H₂

Step 1:- Calculation of Number of moles of Al :-

We know

Number of moles = Mass of substance in g / Gram molar mass

given mass of Al = 0.506 g and

Gram molar mass = 27 g/mol

so

Number of moles of Al = 0.506 g / 27 g/mol = 0.01874 mol

Hence moles of aluminum reacted = 0.01874 mol

Step 2:- Calculation of number of moles of K[Al(OH)₄] formed :-

From the balanced chemical equation it is clear that

2 moles of Al gives = 2 moles of K[Al(OH)₄]

so

0.01874 mol moles of Al gives = 2 mol x 0.01874 mol / 2 mol of K[Al(OH)₄] = 0.01874 mol

Hence required number of moles of K[Al(OH)₄] = 0.01874 mol