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The rock in a lead ore deposit contains 84% PbS by mass. How many kilograms of the rock must be processed to obtain 2.3 kg of Pb?







kg

Sagot :

jonericfogg
Well take 2.3 kg and divide it by .84 to get 2.738 kg then divide the percent of S in the compounds by adding the atomic masses and dividing the atomic mass of S by the atomic mass of the molecule to get 13.40% S by mass
Then you know that Pb is 86.6% by mass so divide 2.738 by .866 to get your answer. Your answer is 3.162 kg of rock is needed
Edufirst

Answer: 3.162 kg


Explanation:


1) Mass percent = (mass of a part / total mass) × 100


2) 84% PbS by mass.


Percent PbS = ( mass of pure PbS / mass of the rock) × 100 = 84%


⇒ mass of the rock = mass of pure PbS × 100 / 84 =


⇒ mass of the rock = mass of pure PbS / 0.84 ← this is fhe formula that you will uise at the end.


2) Determine the mass of pure PbS that contains 2.3 kg of Pb


i) Unit formula: PbS


ii) molar mass of PbS: 207.2 g/mol + 32.065 g/mol = 239.265 g/mol


iii) proportion:


207.2 g of Pb / 239.265 g of PbS = 2,300 g of Pb / x


⇒ x = 2,300 × 239.265 g of PbS / 207.2 = 2,655.93 g of PbS = 2.656 Kg of PbS


4) Use the formula to determine how many kilograms of the rock contain 2.565 kg of pure PbS


mass of the rock = mass of pure PbS / 0.84 = 2.565 / 0.84 = 3.162 kg

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