21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced.
Mass CoCl2 = 10.27 g
moles CoCl2 = 10.27 g/ 129.839 g/mol=0.07910
mass water = 17.40 - 10.27=7.13 g
moles water = 7.13 / 18.02 g/mol=0.396
0.396/ 0.07910=5
CoCl2 * 5 H2O
moles CaF2 = 85.8 g/ 78.0748 g/mol=1.10
moles Ca = 1.10
mass Ca = 1.10 x 40.078 g/mol=44.1 g
V = 44.1 / 1.55 =28.5 mL
Greetings!!
Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol
Molarity: mol/L
First, find mol.
62.5 g x 1mole ÷ 40.31 g = 1.55 mol
then divide mol and the given liters
1.55mol ÷ 1.50 L= 1.03 M