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))

Answer: H2O

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

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

Expert's answer

A. Identifying limiting reagent;

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

Moles of Al_{4}C_{3 } "=\\dfrac{mass}{MM}=\\dfrac{7.034g}{143.959g\/mol}=0.04886mol"

Moles of H_{2}O required to completely react with Al_{4}C_{3}

From the balanced equation, mole ratios of Al_{4}C_{3}: H_{2}O = 1:12;

Moles of H_{2}O 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 H_{2}O: Al(OH)_{3 }=12:4

Since moles of H_{2}O 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 CH**_{4}**, we need to first calculate moles of CH**_{4}** produced**

Thus mole ratios of water: methane = 12:3

Since moles of water (H_{2}O) is 0.30294 mol

Moles of CH_{4 }"=\\dfrac{3}{12}x0.30294mol = 0.075735mol"

mass = moles x MM

= 0.075735mol x 16.06g/mol

= **1.215g**

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