Question

1.How many mL of 0.523 M HBr are needed to dissolve 8.60 g of CaCO3?

2HBr(aq) + CaCO3(s) CaBr2(aq) + H2O(l) + CO2(g)

2.Calculate the number of milliliters of 0.487 M NaOH required to precipitate all of the Mn2+ ions in 101 mL of 0.628 M MnSO4 solution as Mn(OH)2. The equation for the reaction is:

MnSO4(aq) + 2NaOH(aq) Mn(OH)2(s) + Na2SO4(aq)

Answer 1

The answer to your question is:

Step-by-step 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