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5.58 g iron reacted with 3.21 g sulfur. How many grams of iron sulfide were produced ?
We have the following data:
m(Fe) - mass of iron = 5.58 g
m(S) - mass of sufur = 3.21 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 = ?
* to n(Fe)
* to n(S)
The stoichiometric reaction will be in the same proportion (1 : 1), let us see:
1 mol of Fe -------------- 1 mol of FeS
0.1 mol of Fe ------------ 0.1 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 = 0.1 mol
m(FeS) - mass of iron sulfide = ? (in grams)
MM(FeS) - Molar Mass of iron sulfide = 56 + 32 = 88 g/mol
8.8 grams of iron sulfide