How many liters of 3.0 M NaOH solution will react with 2.4 mol H2SO4? (Remember to balance the equation.)

H2SO4 + NaOH → Na2SO4 + H2O

1.2 L
1.6 L
2.4 L
2.8 L

1.6 L is the volume of NaOH that has reacted

Explanation:

The balanced reaction is:

H₂SO₄ + 2NaOH → Na₂SO₄  + 2H₂O

This is a neutralization reaction between a strong acid and a strong base. The products are the correspond salt and water.

We propose this rule of three:

1 mol of sulfuric acid needs 2 mol of NaOH to react to react

Then, 2.4 moles of H₂SO₄  will react with (2.4 . 2) / 1 = 4.8 moles of NaOH

As molarity is 3M, we can determine the volume of our solution

Molarity (M) = mol / volume(L) → Volume(L) = mol / Molarity

Volume(L) = 4.8 mol / 3 M = 1.6 L

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

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explanation: