Question

When carbon disulfide burns in the presence of oxygen, sulfur dioxide and carbon dioxide are produced according to the following equation. cs2 (l) + 3 o2 (g)  co2 (g) + 2 so2 (g)
a. what is the percent yield of sulfur dioxide if the burning of 25.0 g of carbon disulfide produces 40.5 g of sulfur dioxide?

Answer 1

I approached it from mole method let me know if it's correct

Answer 2

Answer:

95.88%

Step-by-step explanation:

1 mole of CS2 gives 2 moles of Sulfur dioxide. Actual mass = 40.5g

nCS2= 25/76

       = 0.33 moles rounded

0.33 moles ->>>>>>>>>>> 0.66 moles of sulfur dioxide (1:2 ratio)

Maxiumum theoretical mass= 0.66 moles x Relative formula mass (64)

= 42.24g.

% yield = Actual mass of product (40.5)/ maximum theoretical mass(42.24g)x100

= 95.88%