Question

Under certain conditions, the substances iron and chlorine combine to form iron(III) chloride. If 31.5 grams of iron and 60.0 grams of chlorine combine to form iron(III) chloride, how many grams of iron(III) chloride must form?

Answer 1

Answer: The mass of iron (III) chloride that must be formed is 91.32 grams

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$     .....(1)

• For iron:

Given mass of iron = 31.5 g

Molar mass of iron = 55.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron}=\frac{31.5g}{55.8g/mol}=0.564mol$

• For chlorine gas:

Given mass of chlorine gas = 60.0 g

Molar mass of chlorine gas = 71 g/mol

Putting values in equation 1, we get:

$\text{Moles of chlorine gas}=\frac{60.0g}{71g/mol}=0.845mol$

The chemical equation for the reaction of titanium and chlorine gas follows:

$2Fe+3Cl_2\rightarrow 2FeCl_3$

By Stoichiometry of the reaction:

3 moles of chlorine gas produces 2 mole of iron (III) chloride.

So, 0.845 moles of chlorine gas will produce = $\frac{2}{3}\times 0.845=0.563moles$ of iron (III) chloride.

Now, calculating the mass of iron (III) chloride from equation 1, we get:

Molar mass of iron (III) chloride = 162.2 g/mol

Moles of iron (III) chloride = 0.563 moles

Putting values in equation 1, we get:

$0.563mol=\frac{\text{Mass of iron (III) chloride}}{162.2g/mol}\\\\\text{Mass of iron (III) chloride}=(0.563mol\times 162.2g/mol)=91.32g$

Hence, the mass of iron (III) chloride that must be formed is 91.32 grams

Answer 2

Answer:

91.5 grams of iron(III)chloride will be formed

Explanation:

Step 1: Data given

Mass of iron = 31.5 grams

atomic mass iron = 55.845 g/mol

Mass of chlorine = 60.0 grams

Molar mass Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe + 3Cl2 → 2FeCl3

Step 3: Calculate Moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 31.5 grams / 55.845 g/mol

Moles iron = 0.564 g/mol

Step 4: Calculate moles Cl2

Moles Cl2 = 60.0 grams / 70.9 g/mol

Moles Cl2 = 0.846 moles

Step 5: Calculate moles FeCl3

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles FeCl3

For 0.564 moles iron we'll have 0.564 moles FeCl3

Step 6: Calculate mass FeCl3

Mass FeCl3 = moles FeCl3 * molar mass FeCl3

Mass FeCl3 = 0.564 moles * 162.2 g/mol

Mass FeCl3 = 91.5 grams

91.5 grams of iron(III)chloride will be formed