Question

Determine how many grams of silver would be produced, if 12.83 x 10^23 atoms of copper react with an excess of silver nitrate. Given the equation: Cu + 2AgNO3 -> Cu(NO3)2 + 2Ag Avogadro's number: 6.02 x 1023

Answer 1

Answer: The amount of silver produced in the given reaction is 459.63 g.

Step-by-step explanation:

According to mole concept:

1 mole of an atom contains
`6.022* 10^(23)` number of atoms.

For the given chemical equation:

`Cu+2AgNO_3\rightarrow Cu(NO_3)_2+2Ag`

By Stoichiometry of the reaction:

1 mole of copper produces 2 moles of silver.

This means that,
`6.022* 10^(23)` number of atoms of copper produces
`2* 6.022* 10^(23)` number of atoms of silver.

So,
`12.83* 10^(23)` number of atoms of copper will produce =
`\frac{2* 6.022* 10^(23)}* 12.83* 10^(23)=25.66* 10^(23)` number of atoms of silver.

We know that:

Mass of 1 mole of silver = 107.87 g

Using mole concept:

If,
`6.022* 10^(23)` number of atoms occupies 107.87 grams of silver atom.

So,
`25.66* 10^(23)` number of atoms will occupy =
`(107.87g)/(6.022* 10^(23))* 25.66* 10^(23)=459.63g`

Hence, the amount of silver produced in the given reaction is 459.63 g.