1. 280 g of CO
2. 16.4 g of Oâ‚‚
3. Â 42 g of Clâ‚‚
Explanation:
Ans 1
Data Given:
moles of Oâ‚‚= 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 Oâ‚‚ then how many moles of CO will react with 5 moles of Oâ‚‚
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 Oâ‚‚
Ans 2
Data Given:
mass of C₃H₈ = 22.4 g
moles of Oâ‚‚= ?
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 Oâ‚‚ = 2(16) = 32 g/mol
So,
    C₃H₈      +     5O₂   >  3 CO₂   +   4H₂O
1 mol (44 g/mol) Â Â Â 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 Oâ‚‚ react with 22.4 grams of ethane
Ans 3
Data Given:
mass of Rubidium Chlorate = 10 g
moles of Oâ‚‚= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
       2 RbClO₃    2 RbCl  +  3Oâ‚‚ Â
         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 Oâ‚‚ = 2(16) = 32 g/mol
So,
    2 RbClO₃        >   2 RbCl   +   3Oâ‚‚ Â
   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 Clâ‚‚= ?
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 Clâ‚‚ = 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 Clâ‚‚ will react with 46 grams of K