Write the combustion equation for propane (C3H8). What is the sum of the coefficients for the reactants and products? Group of answer choices 13 14 10 11 12

13

Explanation:

Let's consider the unbalanced equation for the combustion of propane.

C₃H₈ + O₂ ⇒ CO₂ + H₂O

We will begin balancing H atoms by multiplying H₂O by 4.

C₃H₈ + O₂ ⇒ CO₂ + 4 H₂O

Now, we balance C atoms by multiplying CO₂ by 3.

C₃H₈ + O₂ ⇒ 3 CO₂ + 4 H₂O

Finally, we get the balanced equation multiplying O₂ by 5.

C₃H₈ + 5 O₂ ⇒ 3 CO₂ + 4 H₂O

The sum of the coefficients is 1 + 5 + 3 + 4 = 13.

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g