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Chapter 8: Problem 96

An amount of \(1.0 \times 10^{-3}\) moles of \(\mathrm{Ag}^{+}\) and \(1.0 \times 10^{-3}\) moles of \(\mathrm{CrO}_{4}^{2-}\) reacts together to form solid \(\mathrm{Ag}_{2} \mathrm{CrO}_{4} \cdot\) What is the amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed? \((\mathrm{Ag}=108, \mathrm{Cr}=52)\) (a) \(0.332 \mathrm{~g}\) (b) \(0.166 \mathrm{~g}\) (c) \(332 \mathrm{~g}\) (d) \(166 \mathrm{~g}\)

Short Answer

Expert verified
Answer: (b) 0.166 g

Step by step solution

01

Write the balanced equation

First, write the balanced chemical equation: $$\mathrm{2Ag}^{+}(aq) + \mathrm{CrO}_{4}^{2-}(aq) \rightarrow \mathrm{Ag}_{2} \mathrm{CrO}_{4}(s)$$
02

Identify the limiting reactant

We are given \(\mathrm{1.0 \times 10^{-3} ~moles}\) of both \(\mathrm{Ag}^{+}\) and \(\mathrm{CrO}_{4}^{2-}\). Observe the stoichiometry in the balanced equation: \(2\) moles of \(\mathrm{Ag}^{+}\) react with \(1\) mole of \(\mathrm{CrO}_{4}^{2-}\) to form \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\). Since we have an equal number of moles of both reactants, \(\mathrm{Ag}^{+}\) will be the limiting reactant, as \(2\) moles of \(\mathrm{Ag}^{+}\) are needed for every \(1\) mole of \(\mathrm{CrO}_{4}^{2-}\).
03

Calculate the moles of the product formed

Now, we have to find how many moles of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) are formed. Since the ratio of \(\mathrm{2Ag}^{+}\) to \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) is \(2:1\), the moles of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed will be half the moles of the limiting reactant \(\mathrm{Ag}^{+}\): $$\text{moles of }\mathrm{Ag}_{2} \mathrm{CrO}_{4} = \frac{1}{2} \times \text{moles of limiting reactant} = \frac{1}{2} \times 1.0 \times 10^{-3} \text{ moles} = 5.0 \times 10^{-4} \text{ moles}$$
04

Convert moles of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) to grams

Calculate the molar mass of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) using the given atomic masses: \((2 \times 108) + 52 + (4 \times 16) = 216 + 52 + 64 = 332 \mathrm{~g/mol}\). Next, convert the moles of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed to grams: $$\text{mass} = (\text{moles}) \times (\text{molar mass}) = (5.0 \times 10^{-4}~\text{moles}) \times (332~\text{g/mol}) = 0.166~\text{g}$$
05

Choose the answer

The amount of \(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\) formed is \(0.166 \mathrm{~g}\), so the answer is: (b) \(0.166 \mathrm{~g}\)

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Most popular questions from this chapter

A \(1.50 \mathrm{~g}\) sample of potassium bicarbonate having \(80 \%\) purity is strongly heated. Assuming the impurity to be thermally stable, the loss in weight of the sample, on heating, is (a) \(3.72 \mathrm{~g}\) (b) \(0.72 \mathrm{~g}\) (c) \(0.372 \mathrm{~g}\) (d) \(0.186 \mathrm{~g}\)

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