SAT Chemistry Problem with Solution - Product/Reactant Relations

SAT Chemistry Problem with Solution –
Product and Reactant Relationships

Today I would like to give a solution to yesterday’s SAT Chemistry problem. You can see the original question here: SAT Chemistry Problem – Product and Reactant Relationships

This problem is from the book 320 SAT Chemistry Problems arranged by Topic and Difficulty Level. Here is the problem one more time, followed by the solution.

Level 4 Product and Reactant Relationships

How many moles of water can be produced by the reaction of 4 moles of hydrogen gas with 4 moles of oxygen gas?

A) 2
B) 4
C) 8
D) 10
E) 16

*Solution: According to the stoichiometry, hydrogen is the limiting reactant. Only 2 moles of water will be produced, choice A.

Remarks: (1) The first step for this problem is to write out the chemicals involved and create a balanced equation. This will allow us to determine the relationship between the reactants. It is important to know the coefficients of the reaction because it is very likely that one reactant will be the limiting reagent or limiting reactant. The unbalanced equation will be H₂ + O₂ → H₂O. For a review of balancing equations, see problem 12 in 320 SAT Chemistry Problems. The balanced reaction ends up being 2H₂ + O₂ → 2H₂O.

(2) Even though there is an equal amount of moles of hydrogen and oxygen given, the hydrogen will run out first. This is because the reaction uses twice as much hydrogen as oxygen. We can compare limiting reagent problems to building sandwiches. In this hypothetical, we will say that a sandwich is two slices of bread and three slices of cheese. The stoichiometry would be 2 Bread + 3 Cheese → 1 sandwich.

If you were given 10 slices of bread and 10 slices of cheese and told to make as many sandwiches as possible, you would see that the cheese runs out first. We can define this mathematically by dividing the total amount of each sandwich ‘reactant’ by its coefficient. Looking at the bread ‘reactant’, we get (10 slices of bread)/(2 slices per sandwich) = 5 sandwiches. Looking at the cheese ‘reactant’, we get (10 slices of cheese)/(3 slices per sandwich) = 3 . 3 sandwiches. We would run out of cheese and have leftover bread. This is what it means for a reactant to be limiting. It prevents the use of the other reactant because it runs out.