How many grams of iron are needed to completely consume 27.8 L of chlorine gas according to the following reaction at 25 °C and 1 atm? iron ( s ) + chlorine ( g ) iron(III) chloride ( s ) grams iron

Respuesta :

Answer:

47.99g

Explanation:

The first step st have a balanced chemical reaction equation.

This is given as;

2Fe + 3Cl2 --> 2FeCl3

From the equation,

2 moles of Fe would react with 3 moles of Cl2.

I mole of Fe = 55.8 (3 s.f) (Molar mass of Fe)

2 mole = 2 * 55.8 = 111.6 g

1 mole of gaas occupies 22.4dm3

3 moles = 3 * 22.4 = 67.2 dm3

1dm3 = 1L

This means 116g would consume 67.2L

Xg would consume 27.8L

x * 67.2 = 27.8 * 116

x = 3224.8/67.2

x = 47.99g (2 d.p)