Answer to Question #162468 in General Chemistry for Maddy Brown

Question

Consider the balanced chemical reaction shown below.

1Al4C3(s) + 12H2O(I) —> 4Al(OH)3(s) + 3CH4(g)

In a certain experiment, 7.034 g of Al4C3(s) reacts with 5.453 g H2O(I).

(A) Which is the limiting reactant? (Example: type Al4C3 for Al4C3(s))

 Answer: H2O

(B) How many grams of Al(OH)3(s) form?

(C) How many grams of CH4(g) form?

Answer 1

A. Identifying limiting reagent;

First, we need to find the number of moles of each reactant

Moles of Al₄C₃ $=\dfrac{mass}{MM}=\dfrac{7.034g}{143.959g/mol}=0.04886mol$

Moles of H₂O required to completely react with Al₄C₃

From the balanced equation, mole ratios of Al₄C₃: H₂O = 1:12;

Moles of H₂O required $=\dfrac{12}{1}x0.04886mol = 0.586mol$

But moles of H2O available are $=\dfrac{5.453g}{18g/mol}=0.30294 mol$

Thus moles of water provided is less than the required amount; hence water is the limiting reagent.

B. To calculate the mass of Al(OH)3 produced we need to calculate the moles of it produced.

Moles of Al(OH)₃ produced is calculated as thus;

Mole ratios of H₂O: Al(OH)₃ =12:4

Since moles of H₂O is 0.30294 mol

Therefore, moles of Al(OH)₃ $=\dfrac{4}{12}x0.30294=0.10098mol$

Mass = moles x MM

= 0.10098mol x 78g/mol

= 7.877g of Al(OH)₃

C. To calculate the mass of CH₄, we need to first calculate moles of CH₄ produced

Thus mole ratios of water: methane = 12:3

Since moles of water (H₂O) is 0.30294 mol

Moles of CH₄ $=\dfrac{3}{12}x0.30294mol = 0.075735mol$

mass = moles x MM

= 0.075735mol x 16.06g/mol

= 1.215g