Answer :
[tex]\frac{number of moles}{Volume}[/tex]Answer:
Explanation:
Given parameters:
Concentration of acid = 0.150M
Volume of base = 35.7mL = 0.0357L
Concentration of base = 0.108M
Unknown:
Volume of acid = ?
The balanced equation of the reactions is given as:
Na₂CO₃ + 2HNO₃ → 2NaNO₃ + CO₂ + H₂O
To find the unkown volume of the acid, we work from the known parameters of the base to the unknown volume of the acid.
Solution
Concentraction is given as:
Molarity = [tex]\frac{Number of moles}{Volume}[/tex]
We first find the number of moles of the base used in the reaction. From the number of moles, we can obtain the volume of the acid used.
Number of moles of base = Molarity x volume of base
= 0.0357 x 0.108 = 0.00386moles
From the balanced reaction equation, we know that:
1 mole of the base reacted with 2 moles of the acid
0.00386 moles of the base would completely react with 0.0077moles
From this, we can now obtain the volume of acid used:
Volume of acid used = [tex]\frac{number of moles of acid}{concentration of acid}[/tex]
Volume of acid = [tex]\frac{0.0077}{0.15}[/tex] = 0.0514L = 51.41mL
The volume in mL of 0.150 M HNO₃ that will completely react with the given Na₂CO₃ is 51.4 mL
To determine the volume of HNO₃ that will completely react with the given Na₂CO₃
From the balanced chemical equation,
Na₂CO₃(aq) + 2HNO₃(aq) → 2NaNO₃(aq) + CO₂(g) + H₂O(l)
This means that 1 mole of Na₂CO₃ reacts with 2 moles of HNO₃ to produce 2 moles of NaNO₃, 1 mole of CO₂ and 1 mole of H₂O
Now, we will determine the number of moles of Na₂CO₃ present in the reaction
From the question,
Volume of Na₂CO₃ = 35.7 mL = 0.0357 L
Concentration of Na₂CO₃ = 0.108 M
From the formula
Number of moles = Concentration × Volume
∴ Number of moles of Na₂CO₃ = 0.108M × 0.0357 L
Number of moles of Na₂CO₃ = 0.0038556 moles
Therefore, 0.0038556 moles of Na₂CO₃ reacted in the reaction
According to the balanced equation,
1 mole of Na₂CO₃ reacts will react completely with 2 moles of HNO₃
∴ 0.0038556 moles of Na₂CO₃ will react with 2 × 0.0038556 moles of HNO₃
Number of moles of HNO₃ = 2 × 0.0038556 moles = 0.0077112 moles
Now, we will determine the volume of 0.150 M HNO₃ that will give this number of moles
From
Number of moles = Concentration × Volume
[tex]Volume = \frac{Number of moles }{Concentration}[/tex]
[tex]Volume = \frac{0.0077112}{0.150}[/tex]
Volume = 0.051408 L = 51.408 mL ≅ 51.4 mL
Hence, the volume in mL of of HNO₃ that will completely react with the given Na₂CO₃ is 51.4 mL
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