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

the answer is definitely c.

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.