Here are the correct answers to each of the questions based on the principles of chemistry:
Question 1: A) The mass of the reactants must be equal to the mass of the products. The total number of moles of the reactants must also be equal to the total number of moles of the products.
Answer: "The mass of the reactants must be equal to the mass of the products. The total number of moles of the reactants must also be equal to the total number of moles of the products."
Question 2: A) Divide the mass of the reactant by its molar mass to find the number of moles of the reactant. Use the chemical equation to find the number of moles of the product. Multiply the number of moles of the product by its molar mass to find the mass of the product.
Answer: "Divide the mass of the reactant by its molar mass to find the number of moles of the reactant. Use the chemical equation to find the number of moles of the product. Multiply the number of moles of the product by its molar mass to find the mass of the product."
Question 3: A) The reaction uses 0.04 mol of silver (Ag). To determine the mass of silver sulfide (Ag2S) produced, first calculate the molar mass of Ag2S:
- Molar mass = 2(108 g/mol) + 32 g/mol = 248 g/mol. Since 0.04 mol of Ag is used, this produces:
- 0.02 mol of Ag2S (from the reaction stoichiometry: 2 Ag → 1 Ag2S)
- Mass of Ag2S = (248 g/mol)(0.02 mol) = 4.96 g
Answer: "2(108 g/mol)+32 g/mol=248 g/mol; (248 g/mol)(0.02 mol)=4.96 g"
Question 4: A) To find how many grams of potassium (K) are needed to produce 23.5 grams of potassium oxide (K2O):
- Calculate the molar mass of K2O: \(2(39 g/mol) + 16 g/mol = 94 g/mol\)
- Determine moles of K2O produced from 23.5 grams: \[ \text{Moles of K2O} = \frac{23.5 \text{ g}}{94 \text{ g/mol}} \approx 0.25 \text{ mol} \]
- The stoichiometry from the equation shows 2 moles of K are needed for every 1 mole of K2O: \[ \text{Moles of K needed} = 2 \times 0.25 \text{ mol} = 0.50 \text{ mol} \]
- Calculate grams of K needed: \[ \text{Grams of K} = 0.50 \text{ mol} \times 39 \text{ g/mol} = 19.5 \text{ g} \]
Answer: "19.5 g"
Question 5: A) When 496.5 grams of Pb(NO3)2 reacts with KBr, the total mass of products can be calculated using the Law of Conservation of Mass.
- Molar mass of Pb(NO3)2 = 331 g/mol
- Molar mass of KBr = 119 g/mol
Based on the reaction:
- 1 mole of Pb(NO3)2 reacts with 2 moles of KBr to yield 1 mole of PbBr2 and 2 moles of KNO3.
Calculate moles of Pb(NO3)2: \[ \frac{496.5 \text{ g}}{331 \text{ g/mol}} \approx 1.5 \text{ mol} \] Using the stoichiometry of the reaction:
- Moles of KBr needed: \(2 \times 1.5 \text{ mol} = 3.0 \text{ mol}\)
- Mass of KBr needed: \(3.0 \text{ mol} \times 119 \text{ g/mol} \approx 357 \text{ g}\)
Total mass of reactants: \[ 496.5 \text{ g (Pb(NO3)2)} + 357 \text{ g (KBr)} = 853.5 \text{ g} \] Total mass of products equals total mass of reactants.
Answer: "853.5 g"
Feel free to ask for further clarifications if needed!
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