Answer
a. 2.57 g As is 0.034 moles As.
b. 7.83 g g $S_{8}$ is 0.03 moles $S_{8}$ .
c. 33.8 g $Al_{2}(CO_{3})_3$ is 1.0548 mol $N_{2}H_{4}$, or 1.05 mol $N_{2}H_{4}$, to three significant figures.
d. 227 g $Al_{2}(SO_{4})_3$ s 0.664 mol $Al_{2}(SO_{4})_3$ .
Work Step by Step
Strategy: We have to convert mass to moles. So the conversion factor is 1 mol/Molar mass.
a. Moles in 2.57 g As.
Molar mass of As is 74.92 g/mol
Therefore :
2.57 g As x $\frac{1 mol (As) }{ 74.92 g (As)}$ = 0.0343 mol As
So 2.57 g As is 0.034 moles As (to three significant figures).
b. Moles in 7.83 g $S_{8}$.
Molar mass of $S_{8}$ = 8 x 32.06= 256.48 g/mol
Therefore :
7.83 g $S_{8}$ x $\frac{1 mol (S_{8}) }{ 256.48 g (S_{8})}$ = 0.0305 mol $S_{8}$
So 7.83 g g $S_{8}$ is 0.03 moles $S_{8}$ .
c. Moles in 33.8 g $N_{2}H_{4}$
Molar mass $N_{2}H_{4}$ =2 x 14.0067 + 4 x 1.0079
Molar mass $N_{2}H_{4}$ = 32.045 g/mol
therefore:
33.8 g x $\frac{1 mol(N_{2}H_{4})}{32.045 g (N_{2}H_{4})}$ = 1.0548 mol $N_{2}H_{4}$
So 33.8 g $Al_{2}(CO_{3})_3$ is 1.0548 mol $N_{2}H_{4}$, or 1.05 mol $N_{2}H_{4}$, to three significant figures.
d. Moles in 227 g $Al_{2}(SO_{4})_3$
Molar mass $Al_{2}(SO_{4})_3$ =2 x 26.9815 + 3 x32.065 + 12 x 15.999
Molar mass $Al_{2}(SO_{4})_3$ = 342.146 g/mol
therefore:
227g x $\frac{1 mol (Al_{2}(SO_{4})_3}{ 342.146 g (Al_{2}(SO_{4})_3)}$ = 0.6635 mol $Al_{2}(SO_{4})_3$
So 227 g $Al_{2}(SO_{4})_3$ s 0.664 mol $Al_{2}(SO_{4})_3$ .
