If you want to produce 2.00 kg of ethylene glycol from the reaction of C2H4Cl2 and Na2CO3, what is the minimum amount of C2H4Cl2 that is needed? C2H4Cl2(l) + Na2CO3(s) + H2O(l)C2H6O2(l) + 2NaCl(aq) + CO2(g)

Answer 1

First, let's find the molar masses of the substances involved:
C2H4Cl2 = (2 * 12.01 g/mol) + (4 * 1.01 g/mol) + (2 * 35.45 g/mol) = 98.96 g/mol
C2H6O2 (ethylene glycol) = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (2 * 16.00 g/mol) = 62.07 g/mol

Now, we use stoichiometry to find the amount of C2H4Cl2 needed to produce 2.00 kg (2000 g) of ethylene glycol:
1. Convert the mass of ethylene glycol to moles: moles of ethylene glycol = mass / molar mass
moles of ethylene glycol = 2000 g / 62.07 g/mol ≈ 32.22 moles

2. Use the balanced equation to relate moles of ethylene glycol to moles of C2H4Cl2:
1 mole of C2H4Cl2 → 1 mole of C2H6O2 (ethylene glycol)
32.22 moles of C2H4Cl2 → 32.22 moles of C2H6O2

3. Convert moles of C2H4Cl2 to grams: mass of C2H4Cl2 = moles * molar mass
mass of C2H4Cl2 = 32.22 moles * 98.96 g/mol ≈ 3188.84 g

The minimum amount of C2H4Cl2 needed to produce 2.00 kg of ethylene glycol is approximately 3188.84 grams.