Answer to Question #165961 in General Chemistry for Anthony Ramirez

Question

An iron cube has the volume of 4.76 cm^3. If iron has a density of 7.680 x 10^3 kg/m^3, how many moles are in the cube?

Answer 1

$ρ = 7.680 \times 10^3 \; kg/m^3 \\ = 7.680 \times 10^6 \;g/m^3 \\ = 7.680 \;g/cm^3$

Proportion:

7.680 g – 1 cm³

x – 4.76 cm³

$x = \frac{7.680 \times 4.76}{1} = 36.55 \;g$

M(Fe) = 55.84 g/mol

$n = \frac{m}{M} \\ = \frac{36.55}{55.84} = 0.65 \;mol$

Answer: 0.65 mol