Ask question

Ask question

All categories

3 years ago

6

For below checmical equation (which may or may not be balanced), list the number of each type of atom on each side of the equati

on, and determine if the equation is balanced.
2 C6H14(l) + 19 O2(g) -------> 12 CO2(g) + 14 H2O(g)

1 answer:

Olegator [25]3 years ago

5 0

Answer:

Left hand side:-

Carbon - 12

HYdrogen - 28

Oxygen - 38

Right hand side:-

Carbon - 12

Hydrogen - 28

Oxygen - 38

Since, the number of atoms each side are equal, the reaction is balanced.

Explanation:

The given reaction is:-

$2C_6H_{14}_{(l)}+19O_2_{(g)}\rightarrow 12CO_2_{(g)}+14H_2O_{(g)}$

Left hand side:-

Carbon - 12

HYdrogen - 28

Oxygen - 38

Right hand side:-

Carbon - 12

Hydrogen - 28

Oxygen - 38

<u>Since, the number of atoms each side are equal, the reaction is balanced.</u>

You might be interested in

Which represents the empirical formula for a3B9

Viktor [21]

A3B9 represents a molecular formula. The representation of the empirical formula for this compound is AB3. This is so because the empirical formula is the simplest ratio of the atoms present in the molecule. You get AB3 when you divide the subscripts of A3B9, this is 3 and 9, by the greatest common factor, which is 3. 3/3 = 1 and 9/3 = 3, so the subscripts for the empirical formula are 1 and 3, which is what AB3 represents. <span>Answer: AB3.</span>

3 0

3 years ago

Fill in the blanks to complete each statement about the heating of Earth's surface

Ad libitum [116K]

Earth takes in thermal energy from the Sun in a process called- thermal radiation. The angle between the direction of the energy and the surface it is striking is called the angle of - solar incidence.

6 0

3 years ago

Read 2 more answers

Explain why the product at the negative electrode is not always a metal.

Elena-2011 [213]

Answer:

Whether you get the metal or hydrogen during electrolysis depends on the position of the metal in the reactivity series: the metal will be produced if it is less reactive than hydrogen. hydrogen will be produced if the metal is more reactive than hydrogen.

3 0

3 years ago

When iron reacts with oxygen what substance is produced?

zhannawk [14.2K]

It will produce iron oxide

8 0

3 years ago

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

<u>Part 1 :</u>

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

<u>Part 2 :</u>

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

<u>Part 3 :</u>

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

<u>Part 4 :</u>

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago