Solution Found!
Answer: Calculate each of the following quantities:(a)
Chapter 3, Problem 13P(choose chapter or problem)
Calculate each of the following quantities:
(a) Mass in grams of \(6.44 \times 10^{-2} \mathrm{~mol} \text { of } \mathrm{MnSO}_{4}\)
(b) Moles of compound in 15.8 kg of \(\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}\)
(c) Number of N atoms in 92.6 mg of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\)
Questions & Answers
QUESTION:
Calculate each of the following quantities:
(a) Mass in grams of \(6.44 \times 10^{-2} \mathrm{~mol} \text { of } \mathrm{MnSO}_{4}\)
(b) Moles of compound in 15.8 kg of \(\mathrm{Fe}\left(\mathrm{ClO}_{4}\right)_{3}\)
(c) Number of N atoms in 92.6 mg of \(\mathrm{NH}_{4} \mathrm{NO}_{2}\)
ANSWER:
Step 1 of 3
(a) Mass in grams of \(6.44\times10^{-2}\mathrm{~mol}\text{ of }\mathrm{MnSO}_4\).
The molar mass of \(\mathrm{MnSO}_{4}\) = 151.01 g/mol
Therefore, the mass of \(\mathrm{MnSO}_4=6.44\times10^{-2}\mathrm{~mol}\text{ of }\mathrm{MnSO}_4\times\frac{151.01\mathrm{~g}}{1.0\mathrm{~mol}\mathrm{\ MnSO}}\)
\(=9.725044=9.73\mathrm{~g}\mathrm{MnSO}_4\)
Thus, the mass of \(6.44\times10^{-2}\mathrm{~mol}\) of \mathrm{MnSO}_{4} is \(9.73\ g\)