Question

Determine how many atoms of pure silver will be created when 19.83 x 1023 atoms of copper are used in the following reaction:

Answer 1

Answer:

$\boxed{3.966 \times 10^{24}\text{ atoms of Ag}}$

Explanation:

(a) Balanced equation

Cu + 2AgNO₃ ⟶ Cu(NO₃)₂+ 2Ag

(b) Calculation

You want to convert atoms of Cu to atoms of Ag.

The atomic ratio is ratio is 2 atoms Ag:1 atom Cu

$\text{Atoms of Ag} = 19.83 \times 10^{23}\text{atoms Cu} \times \dfrac{\text{2 atoms Ag}}{\text{1 atom Cu}}\\\\= 3.966 \times 10^{24}\text{ atoms of Ag}\\\\\text{The reaction will produce }\boxed{\mathbf{3.966 \times 10^{24}}\textbf{ atoms of Ag}}$

Answer 2

Answer:

$3.966\times 10^{24}$ pure silver atoms are produced.

Explanation:

$1 mol = N_A=6.022\times 10^{23} mol^{-1}$

Atoms of copper = $19.83\times 10^{23}$

Moles of copper = $\frac{19.83\times 10^{23} }{N_A}$

$=\frac{19.83\times 10^{23} }{6.022\times 10^{23} mol^{-1}}=3.293 mol$

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

According to reaction, 1 mol of copper atoms gives 2 moles of silver atoms.

Then 3.293 moles of moles of copper will give:

$\frac{2}{1}\times 3.293 mol=6.586 mol$

Number of silver atoms in 6.586 moles of silver:

=$6.586 mol\times 6.022\times 10^{23} mol^{-1}=3.966\times 10^{24} atoms$