Question

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

Answer 1

To find the mass of iron needed to react with 27.8 L of chlorine gas, one must use the stoichiometry of the balanced chemical equation and the ideal gas law to calculate the moles of reactants and convert this to mass using the molar mass of iron.

Step-by-step explanation:

Iron and Chlorine Gas Reaction to Produce Iron(III) Chloride

The balanced chemical equation for the reaction between iron and chlorine gas to form iron(III) chloride (FeCl3) is: 2 Fe (s) + 3 Cl2 (g) → 2 FeCl3 (s). To determine the amount of iron required to react with 27.8 L of chlorine gas at 25 °C and 1 atm, we first need to utilize the ideal gas law to find the number of moles of Cl2.

At standard temperature and pressure (STP), 1 mole of an ideal gas occupies 22.4 L. Since the conditions are not at STP, we must use the ideal gas law, which is PV = nRT. The molar mass of iron (Fe) is 55.85 g/mol. By calculating the moles of Cl2, we can then use the stoichiometry of the balanced equation to find the moles of Fe needed, and finally convert moles of Fe to mass using its molar mass.

After performing these stoichiometric calculations, we will arrive at the mass of iron needed to react with the given volume of chlorine gas.

Answer 2

47.99g

Step-by-step 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)