Se combinan 40g de so2 y 25g de o2, determine el porcentaje en masa del exceso con respecto a su masa inicial. so2 + o2 --> so3

p. a.(s=32 ; o=16)

Answer : The percentage mass of excess with respect to initial mass is 60.0 %

Solution : Given,

Mass of = 40 g

Mass of = 25 g

Molar mass of = 64 g/mole

Molar mass of = 32 g/mole

First we have to calculate the moles of and .

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 mole of react with 1 mole of

So, 0.625 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the excess moles of

Excess moles of = Required moles - Given moles

Excess moles of = 0.781 - 0.312 = 0.469 mole

Now we have to calculate the excess mass of

Excess mass of = 15.0 g

Initial mass of = 25 g

Now we have to calculate the percentage mass of excess with respect to initial mass.

Therefore, the percentage mass of excess with respect to initial mass is 60.0 %

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons