Question

List the number of each type of atom on the left side of the equation 2C6H14(l)+19O2(g)→12CO2(g)+14H2O(g) Enter your answers separated by commas (the order of the numbers is the same as the order of the elements on the left side of the equation).

Answer 1

Answer: The number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively

Explanation:

In a chemical equation, the chemical species are termed as reactants or products.

Reactants are defined as the species which react in the reaction and are written on the left side of the reaction arrow.

Products are defined as the species which are produced in the reaction and are written on the right side of the reaction arrow.

For the given chemical equation:

$2C_6H_{14}(l)+19O_2(g)\rightarrow 12CO_2(g)+14H_2O(g)$

On the reactant side:

Number of carbon atoms = (6 × 2) = 12

Number of hydrogen atoms = (14 × 2) = 28

Number of oxygen atoms = (2 × 19) = 38

Hence, the number of carbon, hydrogen and oxygen atoms on the left side of the reaction are 12, 28 and 38 respectively