Question

PLEASE HELP ASAP

Answer 1

Answer:

See explanation

Explanation:

We must first write the equation of the reaction as follows;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Now;

We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles

So;

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

So;

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

We can now decide on the limiting reactant to be C3H8

Therefore;

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

Again;

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Hence;

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

In order to obtain the percentage yield from the reaction, we have;

b) Actual yield = 269.34 g

Theoretical yield = 396 g

Therefore;

% yield = actual yield/theoretical yield × 100/1

Substituting values

% yield = 269.34 g /396 g × 100

% yield = 68%