what volume of 10 percentage (mass/volume) solution of Na2CO3 will be required to neutralise 100mL of HCl sution containing 3.65g of HCl. please answer with all the steps . No silly answers please.
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We write the formulae for the reaction :
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 =53cm³
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 =53cm³
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