What is the mass of a product if the reactants have mass of 44.6 grams (Fe) and 12.8 grams (O2)? A) 31.8 grams
B) 571 grams
C) 57.4 grams
D) 3.48 grams

The mass of the product (  Fe₂O₃ )  will be ; ( C ) 57.4

The mass of the product when Fe reacts with O₂ will be equal to the summation of the masses of the reactants i.e. ∑ M₁ + M₂. and this is because of the  law of conservation of mass

Reaction equation :  Fe + 0₂ = Fe₂O₃

mass of Fe ( M₁ )  = 44.6 g

mass of O₂ ( M₂ ) = 12.8 g

Therefore the mass of Fe₂O₃

= 44.6 g + 12.8 g = 57.4 grams

Hence we can conclude that the mass of a product will be 57.4 grams .

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C) 57.4 grams

Explanation:

According to the law of conservation of mass/matter, mass can neither be created nor destroyed. This means that the amount of matter present in the reactants must be equal to that of the products since no amount can be lost.

Hence, according to this question which states that the mass of reactants are 44.6 grams (Fe) and 12.8 grams (O2), hence the total mass of reactants is 44.6 + 12.8 = 57.4 grams.

If the amount of mass in the reactants is 57.4 grams, then the amount of mass in the product(s) must also be 57.4 grams in conformity to the law of conservation of mass.

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the answer is na (sodium)