Question

To ensure that the air in the shuttle remains free of excess CO2, engineers test the air-purification system. They combine 1.000 × 103 g LiOH with 8.80 × 102 g CO2. The reaction produces 3.25 × 102 g H2O. What is the theoretical yield of this test reaction? Record answer to the nearest whole number.

Answer 1

See explanation below

Step-by-step explanation:

First, we need to write the overall reaction here which is the following:

2LiOH + CO₂ -------> Li₂CO₃ + H₂O

Now that we have the reaction, let's see the theorical yield. This can be calculated using the initial mass of LiOH and CO₂. From here, we can calculate the moles, and then, see which is the limiting reactant to calculate the moles of the products formed, and then, the final mass, which would be the theorical yield.

So, let's calculate the moles of each reactant, using the molar masses.

MM LiOH = 23.95 g/mol

MM CO₂ = 44 g/mol

The number of moles of each reactant would be:

mol LiOH = 1000 g / 23.95 = 41.75 moles

mol CO₂ = 880 / 44 = 20 moles

We have the moles, let's see who is the limiting reactant:

IF: 2 moles LiOH --------> 1 mol CO₂

Then: 41.75 moles -----------> X

X = 41.75 * 1 / 2 = 20.88 moles of CO₂

However we only have 20 moles of CO₂, therefore, the CO₂ is the limiting reactant.

Now that we know this, let's see how many moles of water and Li₂CO₃ will be produced:

moles CO₂ = moles H₂O = moles Li₂CO₃ = 20 moles

So, we can calculate the mass of both of them, using the molar masses:

MM Li₂CO₃ = 73.89 g/mol

m Li₂CO₃ = 20 * 73.89 =1477.8 g

MM H₂O = 18 g/mol

m H₂O = 20 * 18 = 360 g

In other words, the theorical yield of water for this reaction is 360 g and for Li₂CO₃ is 1477.8 g (rounded to 1478 g).

Now, with this we can calculate the percent yield:

% = 325 / 360 * 100

% = 90.28%