Module 10 Module 10 Homework (part of preparation before coming to workshop) Thi

Question

Module 10 Module 10 Homework (part of preparation before coming to workshop) This homework includes exercises on the topics discussed in the white pages of Module 10. I. What is a limiting reactant? a. The reactant that determines how much product is made. b?The reactant that is in excess. c. The product that you can make the most of. d. The amount of reactants that react with each other Ans: 2. You need 2 pieces of bread, 1 tablespoon of peanut butter and 2 tablespoons of jelly to make a peanut butter and jelly sandwich. If you have 10 pieces of bread, 4 tablespoons of peanut butter and 20 tablespoons of jelly How many sandwiches can you make? a. Ans: b. Which is the limiting reactant? Ans: What is in excess? By how much? c. Ans: 3. In the reaction A +B C+ D, if there is more of reactant B available than is required to completely use all of A, then.. a. B is the limiting reactant. b. Both A and B will limit the reaction. c. A is the limiting reactant d. There is no limiting reactant. Ans:

Explanation / Answer

1.

Answer is option (a)

Limiting reactant determines how many moles of product will be formed theoretically.

2.

Since, it requires 1 table spoon of peanut butter, 2 pieces of bread and 2 tablespoon of jelly is required for making one peanut butter butter and jelly sandwich.

So, 4 tablespoon of peanut butter, 8 pieces of bread and 8 tablespoon of jelly is required for making four (4) peanut butter butter and jelly sandwich.

(b) Limiting reactant is tablespoon of peanut butter.

(c) There are
2 pieces of bread and 16 tablespoon of jelly are in excess.

3.

A   +   B   ------->    C   +   D

B is taken in excess. And there are two reactants A and B.

So, limiting reagent is A.

4.

2C8H18     +     25O2     ----->     16CO2     +     18H2O

In the above balanced reaction equation;

2 moles of C8H18 reacts with 25 moles of O2 to produce 16 moles of CO2.

Since, moles of O2 requires more according to the balanced chemical reaction equation. But it has taken 4 moles which is equal to moles of C8H18.

So, O2 is the limiting reactant.

Now,

25 moles of O2 produces 16 moles of CO2.

1 moles of O2 produces (16/25) moles of CO2.

4 moles of O2 produces 4 x(16/25) moles of CO2.

4 moles of O2 produces 2.56 moles of CO2.

Answer is option (C).

5.

NH3     +     O2     ----->     NO     +     H2O

The balanced reaction equation is expressed as

4NH3     +     5O2     ----->     4NO     +     6H2O

In the above balanced reaction equation;

4 moles of NH3 reacts with 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O.

3.25 g of NH3
Molar mass of NH3 = 17 g/mol
So, 17 g of NH3 = 1 mol
or, 1 g of NH3 = (1/17) mol
or, 3.25 g of NH3 = (3.25/17) mol
                               = 0.191 mol
0.191 x (5/4) = 0.23875

3.50 g of O2
Molar mass of O2 = 32 g/mol
So, 32 g of NH3 = 1 mol
or, 1 g of NH3 = (1/32) mol
or, 3.50 g of NH3 = (3.50/32) mol
                               = 0.109 mol
0.109 x (4/5) = 0.0872

So, the moles of O2 is less and moles of NH3 is more. Hence, O2 is the limiting reagent

Again,

Since, 5 moles of O2 produces 4 moles of NO
or, 1 moles of O2 to produce 4/5 moles of NO
or, 0.109 moles of O2 to produce 0.109 (4/5) moles of NO
or, 0.109 moles of O2 to produce 0.0872 moles of NO

Molar mass of NO = 30 g/mol
So, 1 mole of NO = 30 g
or, 0.0872 moles of NO = 0.0872 x 30 g = 2.616 g

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