Na2C2O4+CaCl2 ᐧᐧᐧᐒ CaC2O4 +2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium chloride required to produce maximum calcium oxalate using the above equation.

Na2C2O4+CaCl2 ᐧᐧᐧᐒ CaC2O4 +2NaCl. Given a solution of 1.9g of sodium oxalate in 50g of water at room temperature, calculate the minimum volume of 0.1 M calcium chloride required to produce maximum calcium oxalate using the above equation.

A. 1.40*10^2 dm^2

B. 1.40*10^2 cm^3

C. 1.40*10^-2 dm^-2

D. 1.40*10^-2 cm^3

The correct answer is B. 1.40*10^2 cm^3

Calculation:

Let’s solve it step by step.

Balanced equation
 Na₂C₂O₄ + CaCl₂ᐧᐧᐧᐒ CaC₂O₄ + 2NaCl

The mole ratio of Na₂C₂O₄ : CaCl₂ = 1 : 1.

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1. Moles of sodium oxalate used

Molar mass of Na₂C₂O₄:
(2 * 23) + (2 * 12) + (4 * 16) = 134  g mol^{-1}

Moles of Na2C2O4 = 1.9/134 = 0.0142  mol

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2. Moles of CaCl₂ required

From the equation, moles required are equal: 

Moles of CaCl2 = 0.0142 mol

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3. Volume of 0.1 M CaCl₂ solution

V = n/C = 0.0142/0.1 = 0.142  dm^3

Convert to cm³:
0.142  dm^3 = 1.42 * 10^2  cm^3

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✅ Correct Answer

B. 1.40 * 10^2 cm^3

QUESTION NUMBER 18