Question

In a particular experiment, 25.0 g of CS2 reacted with 40.0 g oxygen to produce 22.4 g of SO2. CS2+O2-->CO2 +SO2 a) What is the percent yield? b) Calculate the mass in grams of the excess reactant.

Answer 1

Yield = `1.06%`
Excess `O_2 = 17.6g`

Explanation:

First, we calculate the theoretical yield from the balanced equation. Then we compare the actual yield to the theoretical one to find the percent yield. Using the actual yield masses we then calculate the excess reactant.

Molecular Masses: `CS_2 = 76` ; `O_2 = 32`
Balanced Equation:
`CS_2 + 2O_2 → CO_2 + SO_2` ; ONE mole of `CS_2` reacts with TWO moles of oxygen to produce ONE mole each of `CO_2` and `SO_2`.

Reactant moles:
`CS_2 = 25.0/76 = 0.329` ; `O_2 = 40.0/32 = 1.25`
From this we can see already that even theoretically, `CS_2` is the limiting reagent.
Theoretical `SO_2` Product moles: `0.329`

Actual Product moles:
`SO_2 = 22.4/64 = 0.350`
`CO_2 = 0.350` (must be the same as the `SO_2`)
Percent Yield: `(0.350/0.329) = 1.06%`

Calculated `O_2` required (despite the obvious experiment error in obtaining MORE product than possible):
`0.350 xx 2 = 0.70` mole. `0.7 xx 32 = 22.4g`

Remaining (excess) reactant `O_2 = 40.0 – 22.4 = 17.6g`