Ethan S. answered • 03/20/20
Johns Hopkins University student tutor for math, science, and spanish
To solve this problem,we need to determine the limiting reagent, either copper (II) nitrate or sodium hydroxide in the production of copper (II) hydroxide. The way that we can do this is through the comparison of the amount of moles of each substance, but in the problem because we assume that there is excess sodium hydroxide, the number of moles of copper (II) hydroxide will be equal to the moles of the starting material. The molar mass of each substance being worked with is as follows:
Molar mass copper (II) nitrate: 187.56 g/mol
Molar mass copper (II) hydroxide: 97.561 g/mol
First, we have to convert the grams of copper (II) nitrate to moles using the molar mass:
2.7 g Cu(NO3)2 * (1 mol Cu(NO3)2 / 187.56 g Cu(NO3)2) = 0.01439539347 mol Cu(NO3)2
We will now assume that the number of moles of Cu(OH)2 that will be produced will be equal to the number of moles of copper (II) nitrate that we calculated because copper (II) nitrate is our limiting reagent and we assume the sodium hydroxide to be in excess. We will now calculate the grams of copper (II) hydroxide as follows:
0.01439539347 mol Cu(NO3)2 * (1 mol Cu(OH)2 / 1 mol Cu(NO3)2) = 0.01439539347 mol Cu(OH)2
0.01439539347 mol Cu(OH)2 * (97.561 g Cu(OH)2 / 1 mol Cu(OH)2) = 1.404428982 g Cu(OH)2
With these conversion calculations, all the units cancelled out to give us our desired product of 1.404428983 g Cu(OH)2, or ~ 1.4 g Cu(OH)2 when considering significant digits.
Final Answer: ~ 1.4 g Cu(OH)2
J.R. S.
03/20/20