Answer
a. 2.86 g C is 0.238 moles C.
b. 7.05 g $Cl_{2}$ is 0.0994 mol $Cl_{2}$
c. 76 g $C_{4}H_{10}$ is 1.309 mol $C_{4}H_{10}$
d. 0.112 mol $Al_{2}(CO_{3})_3$
Work Step by Step
Strategy: We have to convert mass to moles. So the conversion factor is 1 mol/Molar mass.
a. Moles in 2.86 g C.
Molar mass of C is 12.01 g/mol
Therefore :
2.86 g C x $\frac{1 mol C }{ 12.01 g}$ = 0.238 mol C (to three significant figures).
So 2.86 g C is0.238 moles C.
b. Moles in 7.05 g Cl2.
Molar mass of $Cl_{2}$ = 2 x 35.45
Molar mass of $Cl_{2}$ = 70.9 g/mol.
theerfore:
7.02 g x $\frac{1 mol Cl_{2}}{70.9 g}$ = 0.0994 mol $Cl_{2}$
So 7.05 g $Cl_{2}$ is 0.0994 mol $Cl_{2}$.
c. 76 g $C_{4}H_{10}$
Molar mass $C_{4}H_{10}$ = 4 x 12.01 + 10 x 1.00
Molar mass $C_{4}H_{10}$ = 58.04 g/mol
Therefore :
76g $C_{4}H_{10}$ x $\frac{1 mol_{ C_{4}H_{10}} }{ 58.04 g}$ = 1.309 mol
So 76 g $C_{4}H_{10}$ is 1.309 mol $C_{4}H_{10}$
d. Moles in 26.2 g $Al_{2}(CO_{3})_3$
Molar mass $Al_{2}(CO_{3})_3$ =2 x 26.98 + 3 x12.01 + 9 x 15.99
Molar mass $Al_{2}(CO_{3})_3$ = 233.9 g/mol
therefore:
26.2 g x $\frac{1 mol}{233.9 g}$ = 0.112 mol $Al_{2}(CO_{3})_3$
So 26.2 g $Al_{2}(CO_{3})_3$ s 0.112 mol $Al_{2}(CO_{3})_3$.