Answer :
The answer is going to be x=36 g of water because mass on left side = 64.1 g + 159.9 g + x g = 224 + x g
mass on right side = 161.9 g + 98.1 g = 260 g
mass on right side = 161.9 g + 98.1 g = 260 g
Final answer:
To determine the mass of water that reacted, subtract the total mass of the initial reactants (sulfur dioxide and bromine) from the total mass of the products (hydrogen bromide and sulfuric acid), which yields 36.0 g of water.
Explanation:
The question asks for the determination of the mass of water that reacts in the formation of hydrogen bromide (HBr) and sulfuric acid (H2SO4) from the reaction of sulfur dioxide (SO2) with bromine (Br2) and water (H2O). This is a stoichiometry problem based on the law of conservation of mass in a chemical reaction.
The total mass of products (hydrogen bromide and sulfuric acid) is given by adding the mass of hydrogen bromide (161.9 g) to the mass of sulfuric acid (98.1 g), which amounts to 260.0 g. The total initial mass of the reactants is the sum of the mass of sulfur dioxide (64.1 g) and the mass of bromine (159.9 g), totaling 224.0 g. According to the law of conservation of mass, the mass of the products must equal the mass of the reactants plus any additional reactants there might be, which in this case includes water.
To find the mass of water, we subtract the total mass of the initial reactants from the total mass of the products: 260.0 g (products) - 224.0 g (SO2 and Br2 reactants) = 36.0 g of water. This is the mass of water that reacted.
