Question

When 28.6 kg of pure carbon reacts with 88.2 kg of titanium (IV) oxide, 42.8 kg of Titanium metal is produced with carbon monoxide as a by-product. Calculate the percentage of the yield for this reaction.Calculate the amount of excess reactant.

Answer 1

First, we need the reaction involved here:

TiO2 (s) + 2C (s) ===> Ti (s) + 2CO (g) (balanced)

Data here:

1 kg = 1000 g

28.6 kg C => 28600 g C

88.2 kg TiO2 => 88200 g TiO2

42.8 kg Ti => 42800 g Ti (actual yield, important)

By-product: CO

Main product: Ti (% yield is calculated related it)

Molar masses:

Ti = 47.8 g/mol

TiO2 = 79.8 g/mol

C = 12.0 g/mol

Procedure:

Part 1) The limiting and the excess reactant

TiO2 (s) + 2C (s) ===> Ti (s) + 2CO (g)

79.8 g TiO2 ------ 2 x 12.0 g C

88200 g TiO2 ------ X = 26526.31 g C

So, for 88200 g TiO2, 26526.31 g C is needed but we have 28600 g C. According to this, the excess is The C and the limiting would be the TiO2.

Part 2) The theoretical yield

From the reaction, and from the limiting reactant:

79.8 g TiO2 ----- 47.8 g Ti

88200 g TiO2 ------ X = 52831.57 g Ti

Part 3) The % yield Ti

% yield = (actual yield/theoretical yield) x 100

% yield = (42800 g Ti/52931.57 g Ti) x 100

% yield = 80.9 %

Part 4) Calculate the amount of excess reactant.

For 88200 g TiO2, 26526.31 g C is needed but we have 28600 g C.

The amount of excess reactant = 28600 g C - 26526.31 g C = 2073.69 g C = 2.07 kg approx.