Question

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

Answer 1

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