Write a balanced equation for its combustion reaction.

Chemistry

1 answer:

laila [671]3 years ago

5 0

<h3>Answer:</h3>

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

<h3>Explanation:</h3>

The balanced chemical equation for the combustion of the hydrocarbon in question is;

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.

In this case;

Reactant side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
Product side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms

An equation is balanced by putting appropriate coefficients on reactants and products involved in the reaction.
An equation is balanced so as to obey the law of conservation of mass.

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A cylinder contains 3.1 L of oxygen at 300 K and 2.7 atm. The gas is heated, causing a piston in the cylinder to move outward. T

zubka84 [21]

Answer: The moles of gas present in the cylinder is 0.34 moles.

Explanation:

Given: $P_{1}$ = 2.7 atm, $V_{1}$ = 3.1 L, $T_{1}$ = 300 K

$P_{2}$ = ?, $V_{2}$ = 9.4 L, $T_{2}$ = 610 K

Formula used to calculate the final temperature is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{2.7 atm \times 3.1 L}{300 K} = \frac{P_{2} \times 9.4 L}{610 K}\\P_{2} = \frac{5105.7}{2820} atm\\= 1.81 atm$