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3 years ago

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

Chemistry

1 answer:

Gnoma [55]3 years ago

3 0

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.

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