Question

(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3

Answer 1

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

What is stoichiometry of a reaction?

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3