Question 2, (10 points) : A mobile phone case is made from a new Korean-made lig

Question

Question 2, (10 points) : A mobile phone case is made from a new Korean-made lightweight, high strength steel, containing aluminum, nickel, iron and carbon. Each case has a mass of 0.5g, and 10 million are produced each year. Calculate the amount mined for the following circumstances:

a.) Aluminum, makes up 50% of the alloy, and exists as an ore with 30% Al metal

b.)Iron, makes up 48% of the alloy, and exists as an ore with 10% Fe metal

c.)Nickel, makes up 1% of the alloy, and exists as an ore with 1% Ni metal

d.)Limestone (pure CaCO3) is the source of the remainder

e.)Calculate the total CO2 produced, if each gram of ore takes 10g of CO2 to produce.

Explanation / Answer

Total alloy produced in a year=0.5 g *1000,000=0.5*10^6 g

a) Part of Al in the whole mass of alloy mined=50% of 0.5*10^6 g=0.25 *10^6 g

But Al exists as an ore with only 30% of it to be pure Al,so the actual Al ore mined for the purpose must be much more.

The actual Al ore mined be X, then 30% of X=0.25 *10^6 g or 0.30X=0.25 *10^6 g

so X =0.25 *10^6 g/0.30=0.83*10^6 g of Al ore

b)Part of Fe in the whole mass of alloy mined=48% of 0.5*10^6 g=0.24*10^6 g

Pure Fe present in ore mined=10% of the ore mined

So the amount of ore mined=0.24*10^6 g/0.10=2.4*10^6 g of Fe ore

c)Part of Ni in the whole mass of alloy mined=1% of 0.5*10^6 g=0.005*10^6 g

Pure Fe present in ore mined=1% of the ore mined

So the amount of ore mined=0.005*10^6 g/0.01=0.5*10^6 g of Fe ore

d)Remainder part of the alloy=100%-(50%+48%+1%)=1%

carbon content of alloy=1%

carbon content of alloy=1% of total alloy mass=1% of 0.5*10^6 g=0.01*0.5*10^6 g=0.005 *10^6 g

As Carbon is produced from limestone, so the ore mined is limestone.We need to calculate how much mass of limestone contains 0.005 *10^6 g of Carbon.

1 mole of CaCO3 contains 1 mole of Carbon

or, unit molar mass of CaCO3 contains unit molar mass of Carbon

100.09 g of CaCO3 contains 12 g Carbon

Therefore,0.005 *10^6 g of C is contained in 100.09/12 *0.005 *10^6 g of CaCO3=0.042 **10^6 g of limestone

e) 1 g of ore takes 10 g of CO2

0.042 **10^6 g of limestone ore would take=0.042 *10^6 g* 10=0.42 *10^6 g of CO2

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