$560kg{{C}_{2}}{{H}_{4}}\times \frac{90}{100}\times \frac{80}{100}\times \frac{1mol}{28g}\times \frac{1mol{{C}_{2}}{{H}_{4}}{{(OH)}_{2}}}{1mol{{C}_{2}}{{H}_{4}}}=14/4mol{{C}_{2}}{{H}_{4}}{{(OH)}_{2}}$