Question

Ethylene glycol, C2H6O2 (m.w. = 62.07 g/mol), is used as antifreeze for automobile engines. If you want to produce 2.00 kg of ethylene glycol from the reaction of C2H4Cl2 (m.w. = 98.96 g/mol) and Na2CO3 (m.w. = 105.99 g/mol), what is the minimum amount of C2H4Cl2 that is needed?C2H4Cl2(l) + Na2CO3(s) + H2O(l) C2H6O2(l)+ 2NaCl(aq) + CO2(g)3.19x103 g3.19 g1.23x107 g3.42x103 g2.00 g

Answer 1

First let's see the whole reaction:

C2H4Cl2 + Na2CO3 + H2O -> C2H6O2 + 2NaCl + CO2

We want to produce 2kg or 2000grams of C2H6O2. Let's find out how many moles of C2H6O2 that means, based on the 62.07g/mol molar mass

62.07 g = 1 mol

2000g = x moles

x = 32.2 moles of C2H6O2

By seeing our balanced reaction, we can notice that the molar ration of C2H6O2 and C2H4Cl2 is 1:1, which means for each mol of C2H6O2 produced, we will only need 1 mol of C2H4Cl2, therefore:

1 mol C2H6O2 = 1 mol C2H4Cl2

32.2 moles of C2H6O2 = also 32.2 moles of C2H4Cl2

Now we have the number of moles and the molar mass of C2H4Cl2, let's find the final mass:

98.96 g = 1 mol

x grams = 32.2 moles

x = 3.19 * 10^3