Answer:
1. 280 g of CO
2. 16.4 g of O₂
3. 42 g of Cl₂
Explanation:
Ans 1
Data Given:
moles of O₂= 5 moles
mass of CO = ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
2CO + O₂ -----------> 2CO₂
2 mol 1 mol
So if we look at the reaction 2 mole of CO react with 1 mole of O₂ then how many moles of CO will react with 5 moles of O₂
For this apply unity formula
2 mole of CO ≅ 1 mole of O₂
X mole of CO≅ 5 mole of O₂
By Doing cross multiplication
moles of CO = 2 moles x 5 moles / 1 mol
moles of CO = 10 mole
Now calculate mass of 10 moles of CO
Formula used
mass in grams = no. of moles x Molar mass
Molar mass of CO = 12 + 16 = 28 g/mol
Put values in above formula
mass in grams = 10 moles x 28 g/mol
mass in grams = 280 g
So,
280 g of CO will react with 5 moles of O₂
_________________________
Ans 2
Data Given:
mass of C₃H₈ = 22.4 g
moles of O₂= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
C₃H₈ + 5O₂ -----------> 3CO₂ + 4H₂O
1 mol 5 mol
Convert moles to mass
Molar mass of C₃H₈ = 3(12) + 8(1)
Molar mass of C₃H₈ = 36 + 8 = 44 g/mol
and
molar mass of O₂ = 2(16) = 32 g/mol
So,
C₃H₈ + 5O₂ -----------> 3 CO₂ + 4H₂O
1 mol (44 g/mol) 5 mol (32 g/mol)
44 g 160 g
So if we look at the reaction 44 g of C₃H₈ react with 160 g of O₂, then how many grams of O₂ will react with 22.4 g of ethane
For this apply unity formula
44 g of C₃H₈ ≅ 60 g of O₂
grams of O₂ ≅ 22.4 g of ethane
By Doing cross multiplication
grams of O₂ = 22.4 g x 44 g/ 60 g
grams of O₂ = 16.4 g
16.4 g of O₂ react with 22.4 grams of ethane
______________________
Ans 3
Data Given:
mass of Rubidium Chlorate = 10 g
moles of O₂= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
2 RbClO₃ ------------ 2 RbCl + 3O₂
2 mol 3 mol
Convert moles to mass
Molar mass of RbClO₃ = 85.5 + 35.5 + 3(16)
Molar mass of RbClO₃ = 169
and
molar mass of O₂ = 2(16) = 32 g/mol
So,
2 RbClO₃ ------------> 2 RbCl + 3O₂
2 mol ( 169 g/mol) 3 mol (32 g/mol)
338 g 96 g
So if we look at the reaction 338 g of RbClO₃ gives 96 g of O₂, then how many grams of O₂ will be given by 10 g of RbClO₃
For this apply unity formula
338 g of RbClO₃ ≅ 96 g of O₂
grams of O₂ ≅ 10 g of RbClO₃
By Doing cross multiplication
grams of O₂ = 338 g x 10 g/ 96 g
grams of O₂ = 35.2 g
35.2 g of O₂ will be produce by 10 grams of RbClO₃
______________________
Ans 4
Data Given:
mass of K = 46 g
moles of Cl₂= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
2K + Cl₂ ------------> 2KCl
2 mol 1 mol
Convert moles to mass
Molar mass of K = 39 g/mole
and
molar mass of Cl₂ = 2(35.5) = 71 g/mol
So,
2K + Cl₂ ------------> 2KCl
2 mol ( 39 g/mol) 1 mol (71 g/mol)
78 g 71 g
So if we look at the reaction 78 g of K react wit 71 g of Cl₂, then how many grams of Cl₂ will react with 46 g of K
For this apply unity formula
78 g of K ≅ 71 g of Cl₂
46 g of K ≅ X grams of Cl₂
By Doing cross multiplication
grams of Cl₂ = 71 g x 46 g/ 78 g
grams of Cl₂ = 42 g
42 g of Cl₂ will react with 46 grams of K
