(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

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.
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: https://brainly.com/question/16060223
Therefore, 156.114 g of CuNO3