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 the actual amount of SO2 made were (4.3090x10^2) g ehen the theoretical yield was (8.78x10^2) g what was the percent yield?

Enter in scientific notation

Question

Chemistry

Sergio Bernard

9 May, 01:11

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 the actual amount of SO2 made were (4.3090x10^2) g ehen the theoretical yield was (8.78x10^2) g what was the percent yield?

Enter in scientific notation

+3

Answers (1)

Know the Answer?

Litzy Mullins · Answer 1

Percentage yield = 0.49 * 10² %

Explanation:

Given dа ta:

Actual yield of SO₂ = 4.309 * 10² g

Theoretical yield of SO₂ = 8.78 * 10² g

Percentage yield = ?

Solution:

Chemical equation:

2NiS₂ + 5O₂ → 2NiO + 4SO₂

Percentage yield:

Percentage yield = actual yield / theoretical yield * 100

Percentage yield = 4.309 * 10² g / 8.78 * 10² g * 100

Percentage yield = 0.491 * 100

Percentage yield = 49.1%

In scientific notation:

Percentage yield = 0.49 * 10² %