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

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:

+ 3 ⇒ 2 Cl

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.