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?

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 × = 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:

= 1,706 M of HCl

I hope it helps!

yes

explanation:

search chemistry final b exam in google and a site called quizlet will pop up

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"