Question

14. Check the following problem for errors. If you find an error, identify it, tell why it is an error and correct the error solving the problem correctly. Convert 0.45 g zinc hydroxide to moles. (Hint: there are errors)

0.65 g ZnOH 82.41 mol = 3.22 X 10 25 mol
6.02X1023 g

I need this done ASAP!!

Answer 1

Zinc hyrdoxide is Zn(OH)2 and it's molar mass is 99.42 grams/mole. The conversion is done like this:
0.45 g Zn(OH)2 x (1 mole / 99.42 g) = .0045 mole Zn(OH)2
The error is in the mass, the chemical formula, and the molar mass of zinc hydroxide.

Answer 2

1) First error is chemical formula of zinc hydroxide. Correct formula for this compound is Zn(OH)₂, because zinc has oxidation number +2 and hydroxide anion has negative charge -1.

Second error is molar mass of zinc hydroxide.

M(Zn(OH)₂) = Ar(Zn) + 2Ar(O) + 2Ar(H) · g/mol.

M(Zn(OH)₂) = 65.38 + 2 · 16 + 2 · 1.01 · g/mol.

M(Zn(OH)₂) = 99.4 g/mol; molar mass of zinc hydroxide.

2) Third error is the unit, unit for molar mass is g/mol (gram per mole), not mol.

Fourth error is formula for calcutating amount of zinc hydroxide.

It should be done like this:

m(Zn(OH)₂) = 0.45 g; mass of zinc hydroxide.

M(Zn(OH)₂) = 92.4 g/mol.

n(Zn(OH)₂) = m(Zn(OH)₂) ÷ M(Zn(OH)₂).

n(Zn(OH)₂) = 0.45 g ÷ 92.4 g/mol.

n(Zn(OH)₂) = 0.0048 mol; amount of zinc hydroxide.

There is not need to multiply with Avogadro consta (6.022·10²³ 1/mol).