Answer:
The answer to your question is:
1.- volume = 0.151 l or 151 ml
2.- 0.241 l or 241 ml of NaOH
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
