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Get Full Access to Introductory Chemistry - 5 Edition - Chapter 8 - Problem 48p
Get Full Access to Introductory Chemistry - 5 Edition - Chapter 8 - Problem 48p

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# Solved: For the reaction shown, find the limiting reactant for each of the initial

ISBN: 9780321910295 34

## Solution for problem 48P Chapter 8

Introductory Chemistry | 5th Edition

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Problem 48P

For the reaction shown, find the limiting reactant for each of the initial quantities of reactants.

$$4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)$$

(a) $$1 \mathrm{~mol} \mathrm{Cr} ; 1 \mathrm{~mol} \mathrm{O}_{2}$$

(b) $$4 \mathrm{~mol} \mathrm{Cr} ; 2.5 \mathrm{~mol} \mathrm{O}_{2}$$

(c) $$12 \mathrm{~mol} \mathrm{Cr} ; 10 \mathrm{~mol} \mathrm{O}_{2}$$

(d) $$14.8 \mathrm{~mol} \mathrm{Cr} ; 10.3 \mathrm{~mol} \mathrm{O}_{2}$$

Step-by-Step Solution:

Step 1 of 4

a)

The limiting reactant in a reaction is the reactant that produces the least amount of product.

The given chemical reaction is as follows.

$$4 C r(s)+3 O_{2}(g) \rightarrow 2 C r_{2} O_{3}(s)$$

From the balanced chemical reaction, four moles of chromium react with three moles of oxygen molecules to produce two moles of chromium oxide.

Let's calculate the number of moles of product that can be produced from one mol of "Cr" and one mol of oxygen.

\begin{aligned}1 \mathrm{~mol} C r \times \frac{2 \mathrm{~mol} \mathrm{Cr}_{2} \mathrm{O}_{3}}{4 \mathrm{~mol} \mathrm{Cr}}=0.5 \mathrm{~mol} \mathrm{Cr}_{2} \mathrm{O}_{3} \\1 \mathrm{~mol} \mathrm{O}_{2} \times \frac{2 \mathrm{~mol} \mathrm{Cr}_{2} \mathrm{O}_{3}}{3 \mathrm{~mol} \mathrm{O}_{2}}=0.67 \mathrm{~mol} \mathrm{Cr}_{2} \mathrm{O}_{3}\end{aligned}

Hence, "Cr" produces less amount of chromium oxide.

Therefore, the limiting reactant is $$1 \mathrm{~mol} \mathrm{Cr}$$.

Step 2 of 4

Step 3 of 4

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