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.

Explanation:

According to mole concept:

1 mole of an atom contains $6.022\times 10^{23}$ 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, $6.022\times 10^{23}$ number of atoms of copper produces $2\times 6.022\times 10^{23}$ number of atoms of silver.

So, $12.83\times 10^{23}$ number of atoms of copper will produce = $\frac{2\times 6.022\times 10^{23}}\times 12.83\times 10^{23}=25.66\times 10^{23}$ number of atoms of silver.

We know that:

Mass of 1 mole of silver = 107.87 g

Using mole concept:

If, $6.022\times 10^{23}$ number of atoms occupies 107.87 grams of silver atom.

So, $25.66\times 10^{23}$ number of atoms will occupy = $\frac{107.87g}{6.022\times 10^{23}}\times 25.66\times 10^{23}=459.63g$

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