The reaction of fe₂o₃ with co produces fe(s) and co₂(g). what mass of fe₂o₃ is required to produce 5.00 kg of fe(s) if the percent yield for the reaction is 78.5%? you may assume that co(g) is present in excess.

2.75g

Explanation:

Firstly, we write a balanced equation of the reaction.

Fe2O3(s)+3CO(g)-->2Fe(s)+3CO2(g).

Now, we write the formula for the percentage yield.

Percentage yield = Actual yield/theoretical yield * 100%

From the equation, we can see that we have one unknown

We convert the 5kg to g. It must be known that 5kg is 5000g as 1kg equals 1000g.

Now, we are asked to calculate the mass of the iron iii oxide required.

From the balanced equation we can see that 1 mole of iron trioxide yielded 2 moles of Fe.

Now, let’s calculate the mass required. We calculate the actual number of moles of iron produced. This is the mass of iron divided by the molar mass. The atomic mass of iron is 56 amu. The number of moles of iron produced is 5000/56 moles

Since the mole ratio is 1 to 2, the number of moles of iron trioxide thus used is 5000/56 divided by 2 which equals 5000/112 moles.

Now, we proceed to calculate the mass of iron trioxide used. The mass is equal the molar mass multiplied by the number of moles. The molar mass of iron trioxide is 2(56) + 3(16) = 112 + 48 = 160g/mol

The mass thus required to produce 5kg of iron is 5000/112 * 160 which is 3,500g or 3.5kg. We know this to be the theoretical mass, the actual mass is calculated using the formula given above.

78.5 = actual mass/3.5 * 100

Actual mass = (3.5 * 78.5) /100

= 2.75g

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