Answer:
The answer to your question is: 110 g of Na₂CO₃
Explanation:
Data
Na₂CO₃
concentration = 0.73 M
volume = 1.421 l
MW Na₂CO₃ = (2 x 23) + (12 x 1) + (16 x 3) = 106 g
Process
Molarity = moles / volume
moles = Molarity x volume
moles = 0.73 x 1.421
moles = 1.03733
1 mol of Na₂CO₃ --------------------- 106 g
1.03733 ---------------------- x
x = (1.03733 x 106) / 1
x = 109.95 g ≈ 110 g