Question

2. Which of the following is true about the total number of reactants and the total number of

Answer 1

9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

• For example 8O₂ depicts 8 moles of two oxygen atoms.
• The number of moles  is a unit for quantifying particles.
• You can liken it to a dozen, gross or a score.
• Since a mole of a substance contains avogadro number of particles. We can relate the number of moles to other parameters.

Learn more:

Moles brainly.com/question/2272966

#learnwithBrainly

Answer 2

Answer:

A) 9 moles of reactants chemically change into 11 moles of product.

Explanation:

C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)

The first step is to check if the equation is indeed a balanced one before proceeding to compare the moles.

This is done by comparing the number of atoms on the reactant side with that in the product side, if equal the reaction is indeed a balanced one.

Number of moles of Carbon atoms

Reactant = 5

Product = 5

Number of moles of Hydrogen atoms

Reactant = 12

Product = 12

Number of moles of Oxygen atoms

Reactant = 16

Product = 10 + 6 = 16

Since the equation is balanced, we can now compare the moles.

From the reactant side we have a total number of 9 (1 + 8 ) moles reacting to yield a total of 11 ( 5 + 6 ) moles of product.

This alone leads us  to take option A as our answer.

Let us see how other options are wrong however;

Option B) It cannot be grams as what chemical equations shows us the molar (number of moles) relationship between elememts/molecules/compounds/ions etc. The gram relationship can be calculated however if the molar mass of the compounds is taken into consideration.

Option C) This is wrong again due to same reasons mentioned for option B.

Option C) In the compound C5H12 alone, there are 17 (5 C and 12 H) atoms. Hence this option is also wrong.