Answer:
The answer to your question is:
Explanation:
1.-
HBr = 0.523 M V = ?
CaCO3 = 8.6 g
2HBr(aq) + CaCO₃(s) ⇒ CaBr₂(aq) + H₂O(l) + CO₂(g)
MW CaCO₃ = 40 + 12 + 48 = 100 g
MW HBr = 80 + 1 = 81 g
Molarity = moles / volume
100 g of CaCO₃ ---------------- 1 mol
8.6 g ---------------- x
x = (8.6 x 1) / 100
x = 0.086 moles
2 moles of HBr ----------------- 1 mol of CaCO₃
x ----------------- 0.086 moles
x = (0.086 x 2) / 1 = 0.172 moles of HBr
Volume = moles / molarity
Volume = 0.172/ 0.523 = 0.323 l or 323 ml of HBr
2.-
V = ? ml NaOH 0.487 M
V = 101 ml of 0.628 M MnSO₄
MnSO₄(aq) + 2NaOH(aq) ⇒ Mn(OH)₂(s) + Na₂SO₄(aq)
MW MnSO₄ = 55 g
MW NaOH = NaOH = 40 g
Moles = Molarity x volume
Moles = (0.628) x (0.101)
Moles = 0.065 moles of MnSO₄
1 mol of MnSO₄ ------------------ 2 moles of NaOH
0.065 ----------------- x
x = (0.065x 2) / 1
x = 0.131 moles of NaOH
Volume = moles / molarity
Volume = 0.131 / 0.487
Volume = 0.268 l or 268 ml of NaOH
