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3 years ago

How many moles of water can be obtained from the reaction of 4 moles of O2?

Chemistry

1 answer:

Yuliya22 [10]3 years ago

6 0

Answer: 8moles

Explanation:

The reaction below shows the formation of 2 moles of water from 2 moles of hydrogen and 1 mole of oxygen respectively.

2H2(g) + O2 (g) --> 2H2O(l)

So, if 1 mole of O2 produce 2 mole of H2O

4 moles of O2 will produce Z mole of H2O

To get the value of Z, cross multiply

1 x Z = 4 x 2

Z = 8

So, the equation will be 8H2(g) + 4O2 (g) --> 8H2O(l)

Thus, 4 moles of O2 will produce 8moles of H2O .

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Answer:

The coefficients of the balanced equation are:

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Explanation:

The Law of Conservation of Matter, also called the Law of Conservation of Mass or the Lomonósov-Lavoisier Law, postulates that "mass is neither created nor destroyed, it is only transformed". This means that the reagents interact with each other and form new products with different physical and chemical properties than the reagents, but the amount of matter or mass before and after a transformation (chemical reaction) is always the same. In other words, then the mass before the chemical reaction is equal to the mass after the reaction.

Then, you must balance the following chemical equation:

v Cu + w HNO₃ → x Cu(NO₃)₂ + y NO + z H₂O

For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts

Left side: 1 copper, 1 hydrogen, 1 nitrogen and 3 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Generally, hydrogen and oxygen balance in the end. So you balance nitrogen first, because copper is already balanced (there is the same amount on both sides of the reaction). If w = 2, then:

Left side: 1 copper, 2 hydrogen, 2 nitrogen and 6 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

But you see then that the oxygen is unbalanced and you have less quantity in the reagents. Then w must be greater. Being w = 4:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 2 hydrogen, 3 nitrogen and 8 oxygen.

Going back to the idea of balancing nitrogen, being y = 2:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 2 hydrogen, 4 nitrogen and 9 oxygen.

Balancing the hydrogen, being z = 2:

Left side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.
Right side: 1 copper, 4 hydrogen, 4 nitrogen and 10 oxygen.

Since you have the same amount of each element on each side of the reaction, the reaction is balanced. Then, the balanced equation is:

Cu + 4 HNO₃ → Cu(NO₃)₂ + 2 NO + 2 H₂O

Finally:

v= 1
w= 4
x= 1
y= 2
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