How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid) I really couldn't find the answer since molarity and volume for sodium carbonate are not given.
I will mark the correct answer with steps as best answer.

34 mL

Explanation:

We'll begin by calculating the number of mole in 1.25 g of sodium carbonate, Na₂CO₃. This can be obtained as follow:

Mass of Na₂CO₃ = 1.25 g

Molar mass of Na₂CO₃ = (23×2) + 12 + (16×3)

= 46 + 12 + 48

= 106 g/mol

Mole of Na₂CO₃ =?

Mole = mass /molar mass

Mole of Na₂CO₃ = 1.25 / 106

Mole of Na₂CO₃ = 0.012 mole

Next, we shall determine the number of mole HCl needed to react with 0.012 mole of Na₂CO₃.

The equation for the reaction is given below:

Na₂CO₃ + 2HCl —> H₂CO₃ + 2NaCl

From the balanced equation above,

1 mole of Na₂CO₃ reacted with 2 moles of HCl.

Therefore, 0.012 mole of Na₂CO₃ will react with = 0.012 × 2 = 0.024 mole of HCl.

Next, we shall determine the volume of HCl required for the reaction. This is illustrated:

Mole of HCl = 0.024 mole

Molarity of HCl = 0.715 M

Volume of HCl =?

Molarity = mole /Volume

0.715 = 0.024 / volume of HCl

Cross multiply

0.715 × volume of HCl = 0.024

Divide both side by 0.715

Volume of HCl = 0.024 / 0.715

Volume of HCl = 0.034 L

Finally, we shall convert 0.034 L to mL

This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.034 L = 0.034 L × 1000 mL / 1 L

0.034 L = 34 mL

Therefore, 34 mL of HCl is needed for the reaction.

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