Question

Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed. In the laboratory, 10 grams of CCl4 are actually produced. Which is the limiting reactant? What is the theoretical yield of CCl4 in grams? 2. Limiting Reagent: Theoretical Yield: % Percent Yield: xcess (2.5 pts.) (2.5 pts.) (2.5 pts.) (2.5 pts.) reactant remaining

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Answer 1

The molar masses of chloroform, chlorine, carbon tetrachloride and hydrogen chloride are 119.38g/mol, 70.906g/mol, 153.82g/mol and 36.46g/mol respectively. The balanced equation of the reaction is as CHCl₃ + Cl₂ -----> CCl₄ + HCl. From this it can be seen applying law of conservation of mass that 1 mole of the reactants give 1 mole of each of the reactants. So, calculating the no.of moles of chloroform and chlorine gas in 25g, we get 25/119.38 = 0.2094 moles and 25/70.906 = 0.3525 moles respectively. Since there is a lower no.of moles of chloroform available in the reaction mixture, it will limit the reaction by producing only that many moles of product as after it is complete consumption, the other reactant alone cannot proceed further to products. This gives us 0.2094moles of carbon tetrachloride and hydrogen chloride which arrives to 0.2094x153.82 = 32.2099g and 0.2094x36.46 = 7.6347g respectively.

Therefore, the limiting reagent has been found to be chloroform in this case and the theoretical yield of carbon tetrachloride is 32.2099g.

Percentage yield is calculated as (Experimental Yield/Theoretical Yield) x 100 = (10/32.2099)x100 = 31.0464%.

Since we found that 0.2094moles of the reactants will be consumed, this leaves us with 0.3525-0.2094 = 0.1431moles of chlorine gas left unconsumed, which amounts to 0.1431x70.906 = 10.1466g of chlorine gas.