Question

4) 300cm of Hydrogen chloride gas were passed over 7.0g of heated iron fillings until there was no further

change. The reaction vessel was then allowed to cool to room temperature. Use the equation below to
determine the mass of iron that remained at the end of the experiment (molar gas volume = 24000cm"; Fe=56).
Fe(s) + 2HCl(g) → FeCl2(s) + H2(g)
(3 mks)
50
3 of IM colution calcium chloride (Avogadro's​

Answer 1

After reacting iron fillings with hydrogen chloride gas, the remaining mass of iron is calculated to be 6.65 g through the use of stoichiometry and the molar volume of a gas.

Step-by-step explanation:

The question asks us to determine the mass of iron that remained after reacting iron fillings with hydrogen chloride gas. To solve this problem, we use stoichiometry based on the balanced chemical equation:

Fe(s) + 2HCl(g) →
`FeCl_2(s) + H_2(g)`

First, we calculate the moles of HCl gas that reacted using the molar volume of a gas at room temperature (24000 cm3/mol).

300 cm3 HCl x (1 mol HCl / 24000 cm3) = 0.0125 mol HCl

Since the reaction consumes two moles of HCl per mole of Fe, the moles of Fe reacted are half that of HCl.

0.0125 mol HCl ÷ 2 = 0.00625 mol Fe

Now, we convert the moles of Fe to mass using its molar mass (56 g/mol).

0.00625 mol Fe x 56 g/mol = 0.35 g Fe

We subtract the mass of iron that reacted from the initial mass to find the remaining mass.

7.0 g Fe - 0.35 g Fe = 6.65 g Fe

Therefore, the mass of iron that remained at the end of the experiment is 6.65 g.

Answer 2

The mass of iron that remained at the end of the experiment : 6.65 g

Further explanation

Given

V HCl gas = 300 cm³

mass Fe = 7 g(Ar Fe = 56 g/mol)

Reaction

Fe(s) + 2HCl(g) → FeCl2(s) + H2(g)

Required

The mass of iron remained

Solution

molar volume for 1 mol gas = 24000 cm³=24 L, so mol HCl :

`\tt (300)/(24000)=0.0125~mole`

mol ratio Fe : HCl from equation : 1 : 2, so mol Fe =

`\tt (1)/(2)* 0.0125=0.00625`

mass of Iron(reacted) :

`\tt 0.00625* 56=0.35~g`

Mass remained

7 g - 0.35 = 6.65 g