A student reacted NiS2 (MM = 122.83 g/mol) with O2 (MM=32.00 g/mol) to make SO2 (MM=64.07 g/mol) according to this balanced equation

2NiS2+5O2=2NiO+4SO2

If (1.50x10^2) g of NiS2 were reacted with (2.420x10^2) g of O2 what would be the theoretical yield of SO2

Enter in scientific notation

Theoretical yield of SO₂ is 1.56 * 10² g

Explanation:

Given dа ta:

Mass of NiS₂ = 1.50 * 10² g

Mass of O₂ = 2.420 * 10² g

Theoretical yield of SO₂ = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Moles of NiS₂:

Number of moles = mass / molar mass

Number of moles = 1.50 * 10² g / 122.83 g/mol

Number of moles = 1.22 mol

Moles of O₂:

Number of moles = mass / molar mass

Number of moles = 2.420 * 10² g / 32 g/mol

Number of moles = 7.56 mol

Now we compare the moles of SO₂ with NiS₂ and O₂.

NiS₂ : SO₂

2 : 4

1.22 : 4/2*1.22 = 2.44

O₂ : SO₂

5 : 4

7.56 : 4/5*7.56 = 6.05

Moles of SO₂ produced by NiS₂ are less it will limiting reactant.

Mass of SO₂ = moles * molar mass

Mass of SO₂ = 2.44 mol * 64.07 g/mol

Mass of SO₂ = 156.33 g

So theoretical yield of SO₂ is 1.56 * 10² g