Na2C2O4 + CaCl2 → 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 calc...

Answer Details

The given chemical equation indicates that 1 mole of sodium oxalate (Na2C2O4) reacts with 1 mole of calcium chloride (CaCl2) to produce 1 mole of calcium oxalate (CaC2O4) and 2 moles of sodium chloride (NaCl). First, we need to calculate the number of moles of sodium oxalate in the given solution. The molecular weight of Na2C2O4 is (2x23) + (2x12) + (4x16) = 134 g/mol. Therefore, 1.9 g of Na2C2O4 is equal to 1.9/134 = 0.01418 moles. According to the equation, 1 mole of Na2C2O4 reacts with 1 mole of CaCl2 to produce 1 mole of CaC2O4. So, the number of moles of CaC2O4 produced will also be 0.01418 moles. Now, we can use the molarity formula, M = moles/volume (in liters), to calculate the minimum volume of 0.1 M CaC2O4 solution required. Rearranging the formula, we get the equation, volume = moles/M. Substituting the values, volume = 0.01418/0.1 = 0.1418 L = 141.8 mL. Therefore, the minimum volume of 0.1 M calcium oxalate solution required is 141.8 mL, which is equivalent to 1.40 x 10^2 cm3. The correct option is: 1.40 χ 102 dm3.