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Who Wants a Sandwich?
Who Wants a Sandwich?
It is very rare that we have all ingredients in the exact proportions so that there are no leftovers. Lets look at an example to better understand this.
lets look at how many sandwiches each ingredient can make separately first.
6 slices of bread = 3 sandwiches
12 pieces of meat = 4 sandwiches
5 slices of cheese = 5 sanwiches
The ingredient that will run out first is the bread. Therefore the bread is the limiting reactant. The other ingredients will have some leftovers when the bread runs out. These leftover ingredients are called excess reactants.
The final number of products produced are always determined by the limiting reactant. So in this case it is only possible to make 3 complete sandwiches.
Solving limiting reactant problems
Solving limiting reactant problems
The simplest way to solve a limiting reactant problem is to solve the problem for each reactant following the steps for mass-mass stoich problems. Then you should have multiple possible answers. The correct answer if the one that produces the least number of products.
How many grams of carbon dioxide can be produced from 10g of methane and 12g of oxygen?
How many grams of carbon dioxide can be produced from 10g of methane and 12g of oxygen?
We are going to solve this problem twice. Once for each reactant then compare the answers. The limiting reactant is the one that produces the least.
Steps to mass-mass stoich problems:
Identify what information we are given and what we need to find.
given: 10.0 g of CH4 and 12.0 g of O2 , find: grams of CO2
Convert given into moles by dividing by molar mass.
10.0 g of CH4÷ 16.04 g/mol CH4 = 0.623 mol CH4
12.0 g O2 ÷ 32.00 g/mol O2 = 0.350 mol O2
Create a conversion factor from the balanced equation.
We need 1 mole CH4 for each 1 carbon dioxide. → CH4 /CO2
We need 2 mole O2 for each 1 carbon dioxide. → 2O2/CO2
Multiply the given by the conversion factor.
0.625 mol CH4 x CO2/CH4 = 0.623 mol CO2*
0.750 mol O2 x CO2/2O2 = 0.1875 mol CO2*
*Notice that the CO2 needs to be in the denominator to cancel out the CO2 in the numerator.
Convert moles to mass by multiplying by the molar mass.
CH4: 0.623 mol CO2 x 44.01 g/mol = 27.4 g CO2
O2 :0.375 mol CO2 x 44.01 g/mol = 8.25 g CO2
The limiting reactant in the one that produces the least carbon dioxide. So the limiting reactant is O2 even though we started with a greater number of grams of it. In this reaction oxygen is needed in twice the amount as methane. So even though we had more oxygen to start with it runs out first.
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Practice
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