In the manufacturing process of sulfuric acid, sulfur dioxide is reacted with oxygen to produce sulfur trioxide. using the equation, 2so2 (g) + o2 imported asset 2so3 (g), if 192 g of sulfur dioxide is given the opportunity to react with an excess of oxygen to produce 225 g of sulfur trioxide, what is the percent yield of this reaction?

The percent yield of this reaction 93.75%.

Explanation:

if 192 g of sulfur dioxide reacts with excessive of oxygen gas.

Moles of Sulfur dioxide =

According to reaction 2 moles of sulfur dioxide gives 2 moles of sulfur trioxide.

Then 3 mol of sulfur dioxide will give:

of sulfur trioxide

Mass of 3 mole of sulfur trioxide = 3 mol × 80 g/mol = 240 g

Experimental yield of sulfur tiroxide = 240 g

Theoretical yield of sulfur trioxide = 225 g

Percentage yield;

The percent yield of this reaction:

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

the answer for sodium is 0.186

the answer for rubidium is 0.251