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.
