Question

What is the molar concentration of HCl in a solution made by reacting 178.5 mL of 2.461 M HCl with 7.140 g of Na2CO3?

Answer 1

Answer:

1,706 M of HCl

Explanation:

In the reaction:

2 HCl + Na₂CO₃ → 2 NaCl + CO₂ + H₂O

The moles of HCl are:

178,5mL≡0,1785 L×2,461 M =0,4393 moles of HCl

The moles of Na₂CO₃ are:

7,140 g ÷ 105.9888 g/mol = 0,06737 moles of  Na₂CO₃

In a total reaction of 0,06737 moles of  Na₂CO₃ you need:

0,06737 moles of  Na₂CO₃ ×$\frac{2 HCl moles}{1 Na_{2}CO_3}$ = 0,1347 moles of HCl

Thus, the restant moles of HCl are:

0,4393 moles - 0,1347 moles = 0,3046 moles of HCl

Thus, molar concentration of HCl is:

$\frac{0,3046 moles}{0,1785 L}$ = 1,706 M of HCl

I hope it helps!