Im not sure how to do this can someone with these?
(the one before the 3 step problems is a 1-2 step problem)

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

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

i believe it is-

d.) gravity pulled particles of dust and gas together to form planets.