Question

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

Answer 1

Hello!

In a chemical reaction, 28g of iron reacts with 16g of sulfur to produce _________ of iron sulfide.

We have the following data:

m(Fe) - mass of iron = 28 g

m(S) - mass of sufur = 16 g

MM(Fe) - molar mass of iron ≈ 56 g/mol

MM(S) - molar mass of sulfur ≈ 32 g/mol

n(Fe) - number of mol of iron = ?

n(S) - number of mol of sulfur = ?

Solving:

* to n(Fe)

$n_{Fe} = \dfrac{m_{Fe}}{MM_{Fe}}$

$n_{Fe} = \dfrac{28\:\diagup\!\!\!\!\!g}{56\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{Fe} = 0.5\:mol}$

* to n(S)

$n_{S} = \dfrac{m_{S}}{MM_{S}}$

$n_{S} = \dfrac{16\:\diagup\!\!\!\!\!g}{32\:\diagup\!\!\!\!\!g/mol}$

$\boxed{n_{S} = 0.5\:mol}$

The stoichiometric reaction will be in the same proportion (1 : 1), let us see:

$Fe + S \Longrightarrow FeS$

1 mol of Fe -------------- 1 mol of FeS

0.5 mol of Fe ------------ 0.5 mol of FeS

Will the reaction of the iron mass with the mass of sulfur produce how many grams of iron sulfide? We will see:

n(FeS) - number of mol of iron sulfide = 0.5 mol

m(FeS) - mass of iron sulfide = ? (in grams)

MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol

Solving:

$n_{FeS} = \dfrac{m_{FeS}}{MM_{FeS}}$

$m_{FeS} = n_{FeS}*MM_{FeS}$

$m_{FeS} = 0.5\:mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}*88\:\dfrac{g}{mol\!\!\!\!\!\!\!\!\!\!\!\dfrac{\hspace{0.6cm}}{~}}$

$\boxed{\boxed{m_{FeS} = 44\:g}}\:\:\:\:\:\:\bf\blue{\checkmark}\bf\green{\checkmark}\bf\red{\checkmark}$

Answer:

44 grams of iron sulfide

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