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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