How many mL of 0.512 M HCl are needed to dissolve 5.42 g of BaCO3? 2HCl(aq) + BaCO3(s) BaCl2(aq) + H2O(l) + CO2(g)

Question

1. How many mL of 0.512 M HCl are needed to dissolve 5.42 g of BaCO₃?

2HCl(aq) + BaCO₃(s)  BaCl₂(aq) + H₂O(l) + CO₂(g)

2.Calculate the number of milliliters of 0.425 M NaOH required to precipitate all of the Fe³⁺ ions in 143 mL of 0.661 M FeI₃ solution as Fe(OH)₃. The equation for the reaction is:

FeI₃(aq) + 3NaOH(aq) Fe(OH)₃(s) + 3NaI(aq)

Answer 1

So, you have the balanced equation for the reaction

2HCl(aq) + BaCO₃(s) ==> BaCl₂(aq) + H₂O(l) + CO₂(g)

Note that the mol ratio of HCl to BaCO₃ is 2:1, which we will use in the following calculation.

moles HCl needed = 5.42 g BaCO₃ x 1 mol BaCO₃/197 g x 2 mol HCl/mol BaCO₃ = 0.0550 mol HCl

ml HCl needed = 0.0550 mol HCl x 1 L/0.512 mol x 1000 ml/L = 107 ml (3 sig. figs.)

FeI₃(aq) + 3NaOH ==> Fe(OH)₃(s) + 3NaI ... balanced equation

moles FeI₃ present = 143 ml x 1 L/1000 ml x 0.661 mol/L = 0.0945 mol FeI₃

moles NaOH needed = 0.0945 mol FeI3 x 3 mol NaOH/mol FeI₃ = 0.284 mol NaOH needed

ml NaOH needed = 0.284 mol NaOH x 1 L/0.425 mol x 1000 ml/L = 668 ml NaOH needed (3 sig. figs.)