Q1) For each of the reactions, calculate the mass (in grams) of the product form
ID: 877866 • Letter: Q
Question
Q1) For each of the reactions, calculate the mass (in grams) of the product formed when 15.62 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.
A) 2K(s)+Br2(l)2KBr(s) "Br is underlined"
Express your answer using four significant figures.
B)
4Cr(s)+3O2(g)2Cr2O3(s) "O2 is underlined"
Express your answer using four significant figures
C)
2Sr(s)+O2(g)2SrO(s) "Sr is underlined"
Express your answer using four significant figures.
Q2)Because of increasing evidence of damage to the ozone layer, chlorofluorocarbon (CFC) production was banned in 1996. However, there are about 100 million auto air conditioners that still use CFC12 (CF2Cl2).
These air conditioners are recharged from stockpiled supplies of CFC12.
Part A
If each of the 100 million automobiles contains 1.0 kg of CFC12 and leaks 22 % of its CFC12into the atmosphere per year, how much chlorine, in kg, is added to the atmosphere each year due to auto air conditioners? (Assume two significant figures in your calculations.)
Express your answer using two significant figures.
Q3)A particular coal contains 2.55% sulfur by mass. When the coal is burned, it produces SO2emissions which combine with rainwater to produce sulfuric acid.
-
Determine how much sulfuric acid (in metric tons) is produced by the combustion of 1.0 metric ton of this coal. (A metric ton is 1000 kg.)
Express your answer using two significant figures.
Q4)A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess O2 gives 22.1 gof a mixture of CO2 and SO2
Part A
Find the mass of sulfur in the original mixture.
Express your answer to two significant figures and include the appropriate units.
Explanation / Answer
Q1) For each of the reactions, calculate the mass (in grams) of the product formed when 15.62 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.
Mass Br2 = 15.62 g
Mol of Br2 = 15.62 g / molar mass of Br2 = 15.62 g / 159.808 g per mol
= 0.0977 mol Br2
Mol of KBr produced = mol of Br2 * 2mol KBr/ 1 mol Br2
= 0.0977 mol Br2 * 2 mol KBr/ 1 mol K
=0.195 mol KBr
Mass of KBr in g = mol KBr * Molar mass of KBr
= 0.195 mol KBr * 119.002 g /mol
=23.26 g
B)
4Cr(s)+3O2(g)2Cr2O3(s) "O2 is underlined
Mass of O2 = 15.62 g
Mol of O2 = 15.62 g / 31.998 g per mol =0.488 mol
Mole of Cr2O3 produced
= 0.488 mol * 2 mol Cr2O3 / 3 mol O2
= 0.325 mol Cr2O3
Mass of Cr2O3 = 0.325 g * 151.989 g/mol
= 49.46 g
C)
2Sr(s)+O2(g)2SrO(s) "Sr is underlined"
15.62 g Sr
Moles of Sr = 15.62 g / 87.62 g per mol = 0.178 mol
Mol SrO = 0.178 mol * 2mol SrO (s) / 2 mol Sr
= 0.178 mol SrO
Mass of SrO = 0.178 mol * 103.619 g/mol
= 0.00172 g
Q2)Because of increasing evidence of damage to the ozone layer, chlorofluorocarbon (CFC) production was banned in 1996. However, there are about 100 million auto air conditioners that still use CFC12 (CF2Cl2).
These air conditioners are recharged from stockpiled supplies of CFC12.
Part A
If each of the 100 million automobiles contains 1.0 kg of CFC12 and leaks 22 % of its CFC12into the atmosphere per year, how much chlorine, in kg, is added to the atmosphere each year due to auto air conditioners? (Assume two significant figures in your calculations.)
Solution:
We calculate total amount of CFC in 100 million automobiles
1 million = 1 E6
100 million = 1 E8
Mass of CFC for 100 million auto air container
= 1 E 8 auto container * 1 kg / 1 auto air container
= 1E8 kg CFC
Amount of CFC escape = 1 E8 kg * 22/ 100 = 2.20 E7 kg
Answer = 2.2 E7 kg
Q3)A particular coal contains 2.55% sulfur by mass. When the coal is burned, it produces SO2emissions which combine with rainwater to produce sulfuric acid.
Determine how much sulfuric acid (in metric tons) is produced by the combustion of 1.0 metric ton of this coal. (A metric ton is 1000 kg.)
Sulfur produced by 1 metric ton coal
= 1000 kg Coal * 2.25 / 100 = 22.5 kg Sulfur
We calculate moles of sulfur
Mol Sulfur = 22.5 * 1E 3 g mass of S / 32.066 g per mol
= 701.7
We know 1 mol H2SO4 contains 1 mol S
Moles of of H2SO4 produced =mol S * 1mol sulfuric acid / 1 mol S
= 701.7 mol S * 1mol H2SO4 / 1 mol S
= 701.7 mol H2SO4
Mass of H2SO4 = 701.7 mol H2SO4 * 98.0778 g /mol
= 6.9 E4 g H2SO4
Q4)A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess O2 gives 22.1 gof a mixture of CO2 and SO2
Part A
Find the mass of sulfur in the original mixture.
Mass of C + mass of S = 9.0 g
Mass of CO2 + Mass of SO2 = 22.1 g
Here we make two equations
Assume x is for mass of C and y is mass of S
x + y = 9.0 g
and we know
12 g C gives 44 g CO2 and 32 g S gives 64 g SO2
Mass of CO2 = (44 g CO2 / 12 g ) * x
Mass of SO2 = (64 g SO2 / 32 g S ) * y
From problem condition
Mass of CO2 + Mass of SO2 = 22.1 g
(44 g CO2 / 12 g C ) * x + (64 g SO2 / 32 g S ) * y = 22.1 g
We got two equations
We use these two equations to get x and y
x + y = 9.0 …….1
(44 / 12) * x + (64/ 32 ) * y = 22.1 …..2
Lets multiply equation 1 by 44 g and divide by 12 g and subtract equation 2 from resulting equation.
(44 /12 ) * x + (44/ 12 ) * y = 9.0 * 44/12
(44 /12 ) * x +(44/ 12 ) * y= 9.0 * 44/12
- (44 / 12) * x + (64/ 32 ) * y = 22.1
---------------------------------------------------------
1.67 y = 10.9
y = 6.53 g
Mass of S = 6.5 g
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