If we react 5.4g of sodium chloride with an unknown amount of fluorine gas, we produce 4.9g of sodium fluoride and 3.7g chlorine gas. How much fluorine was consumed in the reaction?

4.43 g

Explanation:

The reaction between sodium chloride and flourine gas is given as;

NaCl + F2 --> NaF + Cl2

From the stochiometry of the equation;

1 mol of NaCl reacts eith 1 mol of F2 to form 1 mol of NaF and Cl2

Mass of 1 mol of F2 = 38g

Mass of 1 mol of sodium flouride, NaF = 42g

This means 38g of flourine reacted with NaCl to form 42g of NaF

xg of F2 would form 4.9g of NaF

38 = 42

x = 4.9

x = 4.9 * 38 / 42

x = 4.43 g