Question

1.How many mL of 0.401 M HI are needed to dissolve 5.97 g of BaCO3?

2HI(aq) + BaCO3(s) BaI2(aq) + H2O(l) + CO2(g)
2.Calculate the number of milliliters of 0.757 M NaOH required to precipitate all of the Co2+ ions in 167 mL of 0.548 M CoSO4 solution as Co(OH)2. The equation for the reaction is:
CoSO4(aq) + 2NaOH(aq) Co(OH)2(s) + Na2SO4(aq)

Answer 1

The answer to your question is:

1.- volume = 0.151 l or 151 ml

2.- 0.241 l or 241 ml of NaOH

Step-by-step explanation:

1.-

Data

V = ? HI = 0.401 M

BaCO3 = 5.97 g

2HI(aq) + BaCO3(s) ⇒ BaI2(aq) + H2O(l) + CO2(g)

MW BaCO3 = 137 + 12 + 48 = 197 g

197 g of BaCO3 ----------------- 1 mol

5.97 g ----------------- x

x = (5.97 x 1) /197

x = 0.03 mol of BaCO3

2 moles of HI ---------------- 1 mol of BaCO3

x ---------------- 0.03 mol of BaCO3

x = (0.03 x 2) / 1

x = 0.060 mol of HI

Molarity = moles / volume

volume = moles / molarity

volume = 0.060 / 0.401

volume = 0.151 l or 151 ml

2.-

V = ? NaoH 0.757 M

Co⁺² Volume = 167 ml 0.548 M

CoSO4(aq) + 2NaOH(aq) ⇒ Co(OH)2(s) + Na2SO4(aq)

Moles of Co = Molarity x volume

Moles of Co = 0.548 x 0.167

Moles of Co = 0.092

1 mol of CoSO4 -------------- 2 moles of NaOH

0.092 moles --------------- x

x = (0.092 x 2) /1

x = 0.183 moles of NaOH

Volume of NaOH = moles / molarity

= 0.183 / 0.757

= 0.241 l or 241 ml of NaOH