To determine how much sulfur was used in the reaction, we can apply the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. This means that the mass of the reactants must equal the mass of the products.
In this case, we have:
- Mass of iron (Fe) = 60 grams
- Mass of iron sulfide (FeS) produced = 90 grams
Let the mass of sulfur (S) used in the reaction be \( x \) grams.
According to the conservation of mass:
Mass of reactants = Mass of products
This gives us the equation:
\[ \text{Mass of iron} + \text{Mass of sulfur} = \text{Mass of iron sulfide} \]
Substituting the known values into this equation:
\[ 60 , \text{grams (Fe)} + x , \text{grams (S)} = 90 , \text{grams (FeS)} \]
Now, we can solve for \( x \):
\[ x = 90 , \text{grams} - 60 , \text{grams} \]
\[ x = 30 , \text{grams} \]
Thus, the mass of sulfur used in the chemical reaction is 30 grams, which corresponds to option A.