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:

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j