Question

Lead metal can be extracted from a mineral called galena, which contains 86.6% lead by mass. a particular ore contains 68.5% galena by mass. part a if the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 8.00 cm radius?

Answer 1

The volume of sphere can be calculated using the following formula:

$V=\frac{4}{3}\pi r^{3}$

Here, r is radius of the sphere which is 8 cm. Putting the value,

$V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(8 cm)^{3}=2143.57 cm^{3}$

This is equal to the volume of lead, density of lead is $11.34 g/cm^{3}$ thus, mass of lead can be calculated as follows:

$m=d\times V=11.34 g/cm^{3}\times 2143.57 cm^{3}=2.43\times 10^{4}g$

Let the mass of ore be 1 g, 68.5% of galena is obtained by mass, thus, mass of galena obtained will be 0.685 g.

Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:

m=0.685×0.866=0.5932 g

Therefore, 0.5932 g of lead is obtained from 1 g of ore, if the efficiency is 100%.

For 92.5% efficiency, mass of lead obtained will be:

$m=\frac{92.5}{100}\times 0.5932=0.5487 g$

Thus, 1 g of lead obtain from $\frac{1}{0.5487}=1.822$ grams of ore.

Thus, $2.43\times 10^{4} g$ of lead obtain from:

$2.43\times 10^{4}\times 1.822=4.42\times 10^{4}g=44.2 kg$

Therefore, mass of ore required to make lead sphere is 44.2 kg.

Answer 2

Answer:

44kg

Explanation:

Volume of the lead sphere is $= 4/3*π*(r)3$

= 4/3*π*(8.0)3

= 4/3 * π * ( 512)

= 4/3 * 22/7 * (512)

= 2145.52 cm3

The consists of 68.5% galena while galena consists 86.6% lead.

Only 92.5% of the lead can be extracted.

Assuming X as the amount of ore needed

mass of ore can be determined as:

92.5% of 86.6% of 68.5% of X = 2145.52

0.925 * 0.866 * 0.685*X=2145.52

X=0.54871925/2145.52

$=44kg$]