Question

Moles to particles, grams to moles, and percent yield?

I'm having issues with some chemistry homework. If you can, please tell me how you did the problem.

Given this balanced equation:
2H2S + 3O2 --> 2SO2 + 2H2O

a) how many moles of SO2 can be produced from 2.5 moles of H2S?

b) how many grams of SO2 can be produced from 154.6 grams of O2?

c) How many grams of SO2 can be produced from 25.6 grams of H2S and 25.6 grams of O2?

d) From the amounts of reactants used in (c), the actual yield of SO2 is 21.5 grams. What is the percent yield?

Answer 1

You gots...`2H_2S(g)+3O_2(g)rarr2SO_2(g) + 2H_2O(g)`

Explanation:

Putting in some numbers,

`underbrace(2H_2S(g)+3O_2(g))_"164.2 g"rarrunderbrace(2SO_2(g) + 2H_2O(g))_"164.2 g"`

Are you clear as to how the masses of the REACTANTS and PRODUCTS were calculated? As with any chemical reaction both MASS and CHARGE WERE CONSERVED....

Now for `(a)` you gots `2.5*mol` `H_2S`...and we ASSUME that there is a stoichiometric quantity of dioxygen....i.e. `3.75*mol`. At most we can make a `2.5*mol` `SO_2`...a mass of `2.5*molxx64.07*g*mol^-1=160.2*g`

And for `(b)` we gots `154.6*g` `O_2`.......i.e. `4.83*mol` with respect to DIOXYGEN. At most we can make a `4.83*molxx3/2*mol` `SO_2`...a mass of `7.25*molxx64.07*g*mol^-1=464.3*g`.

And for `(c)` we gots `(25.6*g)/(64.07*g*mol^-1)=0.400*mol` `SO_2`, and `(25.6*g)/(32.00*g*mol^-1)=0.800*mol` `O_2`. At most we can make a `0.800*molxx3/2*mol` `SO_2`...a mass of `1.2*molxx64.07*g*mol^-1=76.9*g`.

For `(d)`, I will let you have a go, and post the answer here....