Lead is found in earth’s crust in several different lead ores. suppose a certain rock is composed of 38.0% pbs (galena), 25.0% pbco3 (cerussite), and 17.4% pbso4 (anglesite). the remainder of the rock is composed of substances that don’t contain lead. how much of this rock (in kg) must be processed to obtain 5.8 metric tons of lead? (a metric ton is 1000 kg.)
express your answer using two significant figures.

For each compound, i.e. galena, cerussite and anglesite calculate the quantity of Pb contained.

For that, you need to use the molecular formula and the atomic weights of the elements.

Galena, PbS

Atomic weights

Pb: 207 uma
S: 32 uma
Molar mass = 207 + 32 = 239 g/mol

Proportion of Pb in the compound = 207/239

% of Pb from galena: 38.0% * [207/239]

Cerussite

Pb CO3

Atomic weights
Pb: 207
C: 12
O: 16

Molar mass = 207 + 12 + 3*16 = 267 g/mol

% of Pb coming from the cerussite

25.0% * [207/267]

Anglesite, PbSO4

Molar mass: 207 + 32 + 4*16 = 303 g/mol

% of Pb coming from anglesite = 17.4% * [207/303]

Total quantity of Pb from the rock (call X the quantity of rock)

[38.0%*207/239 + 25.0%*207/267 + 17.4%*207/303]*X.

And that must be equal to 5.8 metric tons = 5800 kg

The result of the calculations are:

[32.91% + 19.38% + 11.89%]X = 5800 KG

 64.18% X = 5800 KG

0.6418 X = 5800 KG

X = 5800 / 0.6418

X = 9037 KG

Which you have to express with two significan figures, i.e. 9000 kg

D. h2o + h2o > h3o + oh- confirmed.