Question #162468

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))

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

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

1
2021-02-11T04:46:26-0500

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

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