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

1,706 M of HCl

Step-by-step 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 ×
`(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:

`(0,3046 moles)/(0,1785 L)` = 1,706 M of HCl

I hope it helps!