What is the mass (in kg) of air in a square room if the room has walls that are

Question

What is the mass (in kg) of air in a square room if the room has walls that are 9.33 feet high and 9.33 long and the density of air is 1.3 g/L?

What is the theoretical yield (in g of precipitate) when 17.1 mL of a 0.625 M solution of iron(III) chloride is combined with 18.3 mL of a 0.744 M solution of lead(II) nitrate?

What is the experimental yield (in g of precipitate) when 15.3 mL of a 0.599 M solution of barium hydroxide is combined with 16 mL of a 0.632 M solution of aluminum nitrate at a 89.7% yield?

Explanation / Answer

a) 9.33 feet = 30.48 × 9.33 = 284.37cm = 28.44dm

Volume of air = 28.44 dm × 28.44dm × 28.44dm = 23003.2dm3 = 23003.2 litre

Density = 1.3g /Litre

Mass of air = 1.3 × 23003.2 = 29904.2 g = 29.90Kg

b) 3Pb(NO3)2 + 2FeCl3 -------> 3PbCl2 + 2Fe(NO3)3

Theoretical mass

Pb(NO3)2 = 3×331.2g = 993.6g

FeCl3 = 2 × 162.204g = 324.408g

PbCl2 = 3×278.1g = 834.3g

993.6g of Pb(NO3) require 324.408g of FeCl3

Actual mass

[ FeCl3 ] = 0.625M

Volume = 17.1ml

No of mole = (0.625/1000)×17.1ml = 0.01069

Mass of FeCl3 = 0.01069 × 162.204 = 1.7335g

[Pb(NO3)2] = 0.744M

Volume = 18.3ml

No of mole =( 0.744/1000)×18.3 = 0.013615

Mass of Pb(NO3)2 = 0.013615 × 331.2 = 4.509g

993.6g of Pb(NO3)2 require 324.408g of FeCl3

4.509 g of Pb(NO3)2 require( 324.408g/993.6)× 4.509 = 1.472g of FeCl3 but actual mass of FeCl3 is 1.7335g

So, FeCl3 is excess and Pb(NO3)2 is limiting

993.6 g of Pb(NO3)2 give 834.3 g of PbCl2

4.509 g of Pb(NO3)2 give (834.3 / 993.6)×4.509 = 3.786g of PbCl2

So, the theoretical yield of PbCl2 = 3.786g

C) 2Al(NO3)3 + 3Ba(OH)2 -----> 2Al(OH)3 +. 3Ba(NO3)2

Theoretical mass

Al(NO3)3 = 2× 212.996 = 425.992g

Ba(OH)2 = 3×171.34g = 514.02g

Al(OH)3 = 2 × 78g = 156g

Actual mass

Ba(OH)2 = 1.570g

Al(NO3)3 = 2.154g

Ba(OH)2 is limiting

Theoretical yield = (156/514.02)×1.570 = 0.476 g of Al(OH)3

percentage yield = 89.7

Experimental yield = (89.7/100)×0.476=0.427g of Al(OH)3 ppt

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