An amount of \(1.0 \times 10^{-3}\) moles of \(Ag ^{+}\)and \(1.0 \times 10^{-3}\) of \(CrO _{4}^{2-}\) react together to form solid \(Ag _{2} CrO _{4}\). What is the amount of \(Ag _{2} CrO _{4}\) formed. \([ Ag =108, Cr =52, O =16]\).
A. 0.33 g B. 0 .67 g C. \(3.52 g\) D. \(0.166 g\)
Correct Answer: D
Explanation
\(2 Ag ^{+}+ CrO _{4}^{2-} \rightarrow Ag _{2} CrO _{4}\) \(2: 1: 100^{2}\) ill react with \(1 / 2 \times 1.0 \times 10^{3} mol\) \(1.0 \times 10^{-3}\) mole of of \(CrO _{4}^{2-}\) to give \(\frac{1}{2} \times 1.0 \times 10^{-3} mol\) of \(Ag _{4} CrO _{4}=5 \times 10^{4} mol\) of \(Ag _{2} CrO _{4}\) Molar mass of \(Ag _{2} CrO _{4}=(108 \times 2)+52+(16 \times 4)\) \(=(216+52+64) g / mol\) \(=332 g / mol\) Mass of \(\begin{aligned} Ag _{2} CrO _{4} &=5 \times 10^{-4} mol \times 332 g / mol \\ &=0.166 g \end{aligned}\)
