Question
Which of the following laboratory procedures best illustrates the law of conservation of mass?
Calculating the number of atoms in 11 g of Na
Using 250 g of impure Cu to obtain 200 g of pure Cu
Heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants
Burning 2.4 g of Mg in an open crucible to produce 2 g of MgO and unused reactants
Calculating the number of atoms in 11 g of Na
Using 250 g of impure Cu to obtain 200 g of pure Cu
Heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants
Burning 2.4 g of Mg in an open crucible to produce 2 g of MgO and unused reactants
Answers
Answered by
GPT-4o mini
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction; thus, the total mass of the reactants must be equal to the total mass of the products.
Among the options given, the one that best illustrates the law of conservation of mass is:
**Heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants.**
In this procedure, the combined mass of sulfur (S) and iron (Fe) as reactants is 32 g + 56 g = 88 g, which corresponds exactly to the mass of the product, FeS (88 g), indicating that any unused reactants do not affect the total mass balance.
The other options either do not provide a mass balance that matches the reactants to products or involve impurities or unused reactants in a less direct way.
Among the options given, the one that best illustrates the law of conservation of mass is:
**Heating 32 g of S and 56 g of Fe to produce 88 g of FeS and unused reactants.**
In this procedure, the combined mass of sulfur (S) and iron (Fe) as reactants is 32 g + 56 g = 88 g, which corresponds exactly to the mass of the product, FeS (88 g), indicating that any unused reactants do not affect the total mass balance.
The other options either do not provide a mass balance that matches the reactants to products or involve impurities or unused reactants in a less direct way.