Question

For each of the reactions, calculate the mass (in grams) of the product formed when 15.12g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant.A)- 2K(s) + Cl2(g) ? 2KCl(s)B)- 2K(s) + Br2(l) ? 2KBr(s)C)- 4Cr(s) + 3O2(g) ? 2Cr2O3(s)D)- 2Sr(s) +O2(g) ? 2SrO(s)

Answer 1

Answer:

Part A : amount of product (KCl) =  28.88 g

Part B :  amount of product (KBr) =  46.13 g

Part C : amount of product (Cr₂O₃) =  17.3 g

Part D: amount of product (SrO) =  35.76 g

Explanation:

Part A:

Data Given:

Reaction :

                      2K(s) + Cl₂(g) --------> 2KCl

Amount of underline Reactant  (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

                                2K(s) + Cl₂(g) --------> 2KCl

                                2mol    1mol                 2mol

it shows that

2 mole of K give 2 mole of  KCl

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KCl = (39 + 35.5)

Molar mass of KCl = 74.5 g/mol

So the look again to reaction in terms of grams

                                      2K(s)     +    Cl₂(g) --------> 2KCl

                            2mole (39 g/mol)                      2mole (74.5 g/mol)

                                      78 g                                  149 g

Apply the Unity formula

                          78 g of Potassium ≅ 149 g of KCl

Then

                        15.12 g of Potassium ≅ how many g of Product (KCl)

By doing cross multiplication

               X g of Product (KCl) = 149 g of KCl  x 15.12 g of K /  78 g of K

              X g of Product (KCl) = 149 g of KCl  x 15.12 g of K /  78 g of K

              X g of Product (KCl) = 28.88 g

So the amount of product (KCl) =  28.88 g

_________________________________________

Part B:

Data Given:

Reaction :

                      2K(s) + Br₂(g) --------> 2KBr

Amount of underline Reactant  (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

                                2K(s) + Br₂(g) --------> 2KBr

                                2mol    1mol                 2mol

it shows that

2 mole of K give 2 mole of  KBr

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KBr = (39 + 80)

Molar mass of KBr =  119 g/mol

So, look again to reaction in terms of grams

                                      2K(s)     +    Br₂(g) --------> 2KBr

                            2mole (39 g/mol)                      2mole (119 g/mol)

                                      78 g                                  238 g

Apply the Unity formula

                          78 g of Potassium ≅  238 g of KBr

Then

                        15.12 g of Potassium ≅ how many g of Product (KBr)

By doing cross multiplication

               X g of Product (KBr) = 238 g of KBr  x 15.12 g of K /  78 g of K

              X g of Product (KBr) = 238 g of KBr  x 15.12 g of K /  78 g of K

              X g of Product (KBr) = 46.13 g

So the amount of product (KBr) =  46.13 g

__________________________________________

Part C:

Data Given:

Reaction :

                      4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

Amount of underline Reactant  (Cr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.

So, amount of product depend on the amount of Chromium (Cr)

Now Look at Given Reaction:

                                  4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

                                  4mol      3mol                 2mol

it shows that

4 mole of Cr give 2 mole of  Cr₂O₃

if we represent mole in grams

Then

Molar mass of Cr = 52 g/mol

Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)

Molar mass of  2Cr₂O₃ =  304 g/mol

So, look again to reaction in terms of grams

                                     4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

                                 4 mol (52 g/mol)              2 mole (304 g/mol)

                                      208 g                                  608 g

Apply the Unity formula

                          208 g of Chromium ≅  608 g of Cr₂O₃

Then

                        15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)

By doing cross multiplication

        X g of Product (Cr₂O₃) = 238 g of Cr₂O₃  x 15.12 g of Cr /  208 g of Cr

        X g of Product (Cr₂O₃) = 238 g of Cr₂O₃  x 15.12 g of Cr /  208 g of Cr

        X g of Product (Cr₂O₃) = 17.3 g

So the amount of product (Cr₂O₃) =  17.3 g

________________________________________

Part D:

Data Given:

Reaction :

                      2Sr(s) + O₂(g) --------> 2SrO(s)

Amount of underline Reactant  (Sr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.

So, amount of product depend on the amount of Strontium (Sr)

Now Look at Given Reaction:

                                    2Sr(s) + O₂(g) --------> 2SrO(s)

                                    2mol      1mol              2mol

it shows that

2 mole of Sr give 2 mole of SrO

if we represent mole in grams

Then

Molar mass of Sr = 87.6 g/mol

Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)

Molar mass of 2SrO =  207.2 g/mol

So, look again to reaction in terms of grams

                                      2Sr(s) + O₂(g) --------> 2SrO(s)

                                 2 mol ( 87.6 g/mol)         2 mole (207.2 g/mol)

                                      175.2 g                                  414.4 g

Apply the Unity formula

                          175.2 g of Strontium ≅  414.4 g of SrO

Then

                        15.12 g of Strontium ≅ how many g of Product (SrO)

By doing cross multiplication

       X g of Product (SrO) = 414.4 g of SrO  x 15.12 g of Sr /  175.2 g of Sr

        X g of Product (SrO) = 414.4 g of SrO  x 15.12 g of Sr /  175.2 g of Sr

        X g of Product (SrO) = 35.76 g

So the amount of product (SrO) =  35.76 g