Answer:
AgNO3 (22.5 g) + BaSO4 (11.7 g) →
Ag2SO4 (41.23 g) + Ba(NO3)2 (13.72 g)
steps
Chemical Reaction Product Mass
Synopsis: In a chemical reaction, 22.5 grams of silver nitrate reacts with 11.7 grams of barium sulfate to form products. We need to find the mass of the products.
Explanation: In the chemical reaction between silver nitrate (AgNO3) and barium sulfate (BaSO4), they react to form products. We are given the mass of both reactants, which are 22.5 grams of AgNO3 and 11.7 grams of BaSO4. To find the mass of the products, we need to balance the chemical equation and determine the molar ratios between the reactants and products.
Balanced Chemical Equation: AgNO3 + BaSO4 → Ag2SO4 + Ba(NO3)2
From the balanced chemical equation, we can see that the molar ratio between AgNO3 and Ag2SO4 is 1:1, and the molar ratio between BaSO4 and Ba(NO3)2 is also 1:1. This means that for every 1 mole of AgNO3 and 1 mole of BaSO4, we will get 1 mole of Ag2SO4 and 1 mole of Ba(NO3)2.
Next, we need to convert the given masses of AgNO3 and BaSO4 to moles using their respective molar masses:
Molar mass of AgNO3 = 169.87 g/mol
Molar mass of BaSO4 = 233.38 g/mol
Number of moles of AgNO3 = 22.5 g / 169.87 g/mol = 0.1324 mol
Number of moles of BaSO4 = 11.7 g / 233.38 g/mol = 0.0501 mol
Since the molar ratios between the reactants and products are 1:1, the number of moles of Ag2SO4 and Ba(NO3)2 formed will also be 0.1324 mol and 0.0501 mol, respectively.
Finally, we can calculate the mass of the products by multiplying the number of moles by their respective molar masses:
Mass of Ag2SO4 = 0.1324 mol x 311.8 g/mol = 41.23 g
Mass of Ba(NO3)2 = 0.0501 mol x 261.34 g/mol = 13.72 g
Therefore, the mass of the products is 41.23 g + 13.72 g = 54.95 g.
Analogy: It's like baking a cake where you need a certain amount of flour and sugar to make the cake, and once you have mixed the ingredients and baked it, you get a cake that weighs a certain amount.
Summary: 22.5 grams of silver nitrate and 11.7 grams of barium sulfate react to form 41.23 grams of silver sulfate and 13.72 grams of barium nitrate.
AgNO3 (22.5 g) + BaSO4 (11.7 g) → Ag2SO4 (41.23 g) + Ba(NO3)2 (13.72 g)
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