Question

Part A (7 points): How many grams of iron (Fe) will be produced if you start with 3.65 liters of carbon monoxide gas (CO) at STP? Part B (3 points): Jacob performed the above experiment in his chemistry class, and collected 4.23 grams of iron. What is the percent yield? Fe2O3 + 3 CO --> 2 Fe + 3 CO2 Please show work.

Please help, this was an optional thing for one of my classes but I really need the extra points.

Answer 1

A. The answer is 6.064 grams of iron.

Since density is mass per unit volume:
     density = mass / volume

we can calculate for the mass of carbon monoxide using the listed density of carbon monoxide at STP which is 1.250 g/L:
     weight = density * volume 
                =1.250 g/L * 3.65 L 
                = 4.5625 grams CO

Using the following values:
     Molar mass of iron is 55.85 g/mol Fe
     Molar mass of carbon monoxide is 28.01 g/mol CO

and taking note of the coefficients of CO and Fe from the reaction
     Fe2O3 + 3 CO --> 2 Fe + 3 CO2

we can solve for the theoretical mass of iron:
     mass of iron = 4.5625 g CO * (1 mol CO / 28.01 g CO) * (2 mol Fe / 3 mol CO) *
                             (55.845 g Fe / 1 mol Fe) 
                          = 6.064 grams Fe

B.  Since Jacob collected 4.23 grams of iron in the above experiment, this is our actual or experimental yield. 

Therefore, the percent yield is 
     Percent yield  = Actual yield / Theoretical yield x 100
                            = (4.23 g Fe / 6.064 g Fe) * 100
                            = 69.76 %