Calculate the mass in grams of 8.694 moles of Ag2CO3.
Correct answer: B. 2399.544 g
- A. 1417.53 g
- B. 2399.544 g
- C. 3456.78 g
- D. 1231.98 g
Explanation
We will use the formula, moles = mass / molecular massMolar mass of Ag2CO3= 275g/molNumber of moles = 8.694 molMass = moles x molar massMass = 8.694 x 275 = 2390g
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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