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Cu + 2NaOH ---> Cu(OH)2 + 2NaSince we have a finite amount of two reactants we must first determine which is the limiting reactant.

For Cu: n= m/M = 0.500 g/63.53 g/mol = 7.87E-3 mol

For NaOH: n = CV = (3.0 mol/L)(0.0300 L) = 9.00E-2 mol

Copper is by far the limiting reactant therefore we will use its amount to find the maximum Cu(OH)2 that can be yielded from the reaction. Since the above reaction scheme indicates that Cu and Cu(OH)2 are in a 1 to 1 ratio, their molar amounts are the same ie. at the end of the reaction there will be 7.87E-3 mol of Cu(OH)2

To find the mass that corresponds to this molar amount we multiply by the molar mass of Cu(OH)2 (97.56 g/mol):

m=nM = (7.87E-3 mol)(97.56 g/mol) = 0.7678 g

Therefore 0.500 g of Cu will yield 0.768 g (3 sig fig) of Cu(OH)2.

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8y ago
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9y ago

Copper (as a metal) doesn't react with sodium hydroxide.

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Q: How many grams of Cu(OH)2 can be produced from the reaction of 0.500g Cu with 30.0mL of 3.0M NaOH?
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