Middle School

If you have 112 g of Ni and 112 g of AgNO₃, how many atoms of silver would be produced?

Answer :

Final answer:

To calculate the number of silver atoms produced from 112 g of AgNO3, we use the molar mass of AgNO3 and Avogadro's number to find that 0.659 moles of AgNO3 yield approximately 0.659 moles of silver, which equals a large number of silver atoms when multiplied by 6.022 x 10^23 atoms/mol.

Explanation:

To determine the number of silver atoms produced from 112 g of AgNO3 when reacted with nickel (Ni), we first need to understand the stoichiometry of the reaction. For each mole of AgNO3, one mole of silver atoms is produced. To calculate the number of moles of AgNO3 we have, use the formula:

Number of moles = mass (g) / molar mass (g/mol)

The molar mass of AgNO3 is approximately 169.87 g/mol. Thus, the number of moles of AgNO3 is:

Number of moles of AgNO3 = 112 g / 169.87 g/mol

This gives us around 0.659 moles of AgNO3. Since the reaction produces one mole of silver for each mole of AgNO3, we will also have 0.659 moles of silver.

To find the number of atoms of silver produced, we use Avogadro's number, which is 6.022 x 1023 atoms/mol:

Number of silver atoms = moles of silver x Avogadro's number

So, number of silver atoms = 0.659 moles x 6.022 x 1023 atoms/mol

This calculation will give us the total number of atoms of silver that can be produced from 112 g of AgNO3.

The number of atoms of silver produced is [tex]\( 3.97 \times 10^{23} \)[/tex] atoms.

To determine the number of atoms of silver produced in the reaction between nickel (Ni) and silver nitrate [tex](AgNO_3)[/tex],

The reaction between nickel and silver nitrate is as follows:

[tex]\[\text{Ni} (s) + 2 \text{AgNO}_3 (aq) \rightarrow 2 \text{Ag} (s) + \text{Ni(NO}_3)_2 (aq)\][/tex]

First, we need to calculate the moles of each reactant.

Moles of Ni:

[tex]\[\text{Molar mass of Ni} = 58.69 \text{ g/mol}\][/tex]

[tex]\[\text{Moles of Ni} = \frac{112 \text{ g}}{58.69 \text{ g/mol}} \approx 1.91 \text{ mol}\][/tex]

Moles of [tex]AgNO_3:[/tex]

[tex]\[\text{Molar mass of AgNO}_3 = 107.87 \text{ g/mol (Ag)} + 14.01 \text{ g/mol (N)} + 3 \times 16.00 \text{ g/mol (O)} = 169.87 \text{ g/mol}\][/tex]

[tex]\[\text{Moles of AgNO}_3 = \frac{112 \text{ g}}{169.87 \text{ g/mol}} \approx 0.66 \text{ mol}\][/tex]

Calculate the Moles of Silver Produced

From the balanced equation, 1 mole of Ni reacts with 2 moles of [tex]AgNO_3[/tex] to produce 2 moles of Ag.

Since we have 1.91 moles of Ni and only 0.66 moles of [tex]AgNO_3[/tex], [tex]AgNO_3[/tex] is the limiting reactant.

According to the stoichiometry of the reaction:

[tex]\[2 \text{ moles of AgNO}_3 \rightarrow 2 \text{ moles of Ag}\][/tex]

Therefore, 0.66 moles of [tex]AgNO_3[/tex] will produce 0.66 moles of Ag.

Convert Moles of Silver to Number of Atoms

We use Avogadro's number to convert moles of silver to atoms.

[tex]\[1 \text{ mole of Ag} = 6.022 \times 10^{23} \text{ atoms of Ag}\][/tex]

[tex]\[\text{Number of atoms of Ag} = 0.66 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole}\][/tex]

[tex]\[\text{Number of atoms of Ag} \approx 3.97 \times 10^{23} \text{ atoms}\][/tex]