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

3.97 * 10²³ atoms of Ag

Explanation:

Given dа ta:

Mass of Ni = 112 g

Mass of AgNO₃ = 112 g

Atoms of silver produced = ?

Solution:

Chemical equation:

Ni + 2AgNO₃ → 2Ag + Ni (NO₃) ₂

Number of moles of Ni:

Number of moles = mass / molar mass

Number of moles = 112 g/58.7 g/mol

Number of moles = 2 mol

Number of moles of AgNO₃:

Number of moles = mass / molar mass

Number of moles = 112 g/169.87 g/mol

Number of moles = 0.66 mol

Now we will compare the moles of Ag with both reactants.

AgNO₃ : Ag

2 : 2

0.66 : 0.66

Ni : Ag

1 : 2

2; 2*2 = 4 mol

Number of moles of Ag produce by AgNO₃ are less so it will limiting reactant and limit the yield of Ag.

I mole = 6.022 * 10²³ atoms

0.66 mol * 6.022 * 10²³ atoms / 1 mol

3.97 * 10²³ atoms of Ag