A reaction has a theoretical yield of 124.3 g SF6, but only 113.7 g SF6 are obtained in the lab, what is the percent yield of SF6 for this reaction?

Percent yield = 91%

Explanation:

Given data:

Theoretical yield of SF₆ = 124.3 G

Actual yield of SF₆ = 113.7 g

Percent yield of SF₆ = ?

Solution:

Formula:

Percent yield = (actual yield / theoretical yield)× 100

By putting values,

Percent yield = (113.7 g/ 124.3 g) × 100

Percent yield = 0.91 × 100

Percent yield = 91%

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.