Asked by Susan
2NaOH + CuSO4 = Na2SO4 + Cu(OH)2
100g of Cu(OH)2 is needed. Assuming 100% yield, how much of 1M NaOH and 0.5M CuSO4 must be used in mL?
Any help would be awesome- even a formula on how to solve
100g of Cu(OH)2 is needed. Assuming 100% yield, how much of 1M NaOH and 0.5M CuSO4 must be used in mL?
Any help would be awesome- even a formula on how to solve
Answers
Answered by
DrBob222
Two stoichiometry problems; one for Cu(OH)2 to CuSO4 and one for Cu(OH)2 to NaOH. Stoichiometry problems, for the most part, are worked the same way.
Step 1. Convert what you have (100 g Cu(OH)2 to mols. mols = grams/molar mass = ?
Step 2. Convert mols Cu(OH)2 to mols NaOH. You can tell from the coefficients in the balanced equation, that mols NaOH = twice the mols Cu(OH)2.
Step 3. Convert mols NaOH to grams. g = mols x molar mass. Assuming 100% yield that is g NaOH needed to produce 100 g of the product you want.
Do the same thing for converting 100 g Cu(OH)2 to g CuSO4.
Step 1. Convert what you have (100 g Cu(OH)2 to mols. mols = grams/molar mass = ?
Step 2. Convert mols Cu(OH)2 to mols NaOH. You can tell from the coefficients in the balanced equation, that mols NaOH = twice the mols Cu(OH)2.
Step 3. Convert mols NaOH to grams. g = mols x molar mass. Assuming 100% yield that is g NaOH needed to produce 100 g of the product you want.
Do the same thing for converting 100 g Cu(OH)2 to g CuSO4.
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