Question

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What volume of 10% (w/v) solution of Na2CO3 will be required to neutralize 100 mL of HCl solution containing 3.65g of HCl.
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Answer 1

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Answer 2

Explanation:

Answer is here

Na₂CO₃(aq) + 2HCl (aq) - - - - > 2NaCl(aq) + CO₂(g) + H₂O (l)

Mole ratio is :

1 : 2 meaning 1 mole of Sodium carbonate reacts with 2 moles of Hydrochloric acid.

Given the mass of HCl we can get the reacting moles.

Moles = mass of solute / molar mass

Molar mass of HCl is : 35.5g (Cl) + 1g (H) =36.5g/mol

Moles = 3.65/36.5 =0.1moles.

The number of reacting moles of Sodium Carbonate is 0.1/2 = 0.05moles from the mole ratio.

The molar mass of Sodium Carbonate is :

23 × 2(Sodium) + 16 × 3(Oxygen) + 12 (Carbon) =106g /mol

Reacting mass of Sodium Carbonate = 106 × 0.05 = 5. 3 g

Given the density as 10 %=0.1g/cm³

Volume = mass / density

5. 3 / 0.1 =53cm^3