When 4.0 mol of CCl4 reacts with an excess of HF, 3.0 mol of CCl2F2 (Freon) is obtained....

When 4.0 mol of CCl₄ reacts with an excess of HF, 3.0 mol of CCl₂F₂ (Freon) is obtained. The equation for the reaction is

State which of the statements are true about the reaction and make the false statements true.

(a) The theoretical yield for CCl₂F₂ is 3.0 mol.

(b) The theoretical yield for HCl is 71 g.

(c) The percent yield for the reaction is 75%.

(d) The theoretical yield cannot be determined unless the exact amount of HF is given.

(e) From just the information given above, it is impossible to calculate how much HF is unreacted.

(f) For this reaction, as well as for any other reaction, the total number of moles of reactants is equal to the total number of moles of product.

(g) Half a mole of HF is consumed for every mole of CCl₄ used.

(h) At the end of the reaction, no CCl₄ is theoretically left unreacted.