Question

7. How many grams of CIF3 form from 130.0 grams of Cl2 when F2 is in excess?

Answer 1

339 grams of CLF3 is formed when F2 is in excess and 130 grams of CL2 reacts.

Explanation:

The balanced chemical reaction for the formation of ClF3 is given by:

$Cl_{2}$ + 3 $F_{2}$ ⇒ 2 Cl$F_{3}$

the mass of Cl2 is given 130 grams

From the equation it is found that 2 moles of chloride reacts to form 2 moles of ClF3.

calculating the number of moles of chlorine by the formula:

Number of moles = mass of the substance ÷ atomic mass of one mole of the substance

n = 130 ÷ 35.45

   = 3.6671 moles

So, applying stoichiometry

2 moles of Cl2 formed 2 moles of ClF3

3.6671 moles of Cl2 will form x moles of ClF3

2 ÷ 2 = x ÷ 3.6671

x = 3.6671 moles of ClF3

now from the formula of number of moles

weight is calculated as n × mass of the gas

                                    3.6671 × 92.448

                                    =  339.01 grams of ClF3 is formed.