Answer:
At the end of the reaction are left 1.20 g of excess reactant (arsenic).
Step-by-step explanation:
1st) The first step to solve this problem is to write the chemical reaction between gallium and arsenic, which is as follow:
Ga + As → GaAs
The reaction is balanced and shows that 1 mol of gallium reacts with 1 mol of arsenic to produce 1 mol of gallium arsenide.
2nd) Now, to find the limiting and excess reactants we need to calculate the number of moles that are included in 4.00 g of gallium and 5.50 g of arsenic. So, let's look for the molar mass of each element in the periodic table:
Gallium molar mass: 69.72 g/mol
Arsenic molar mass: 74.92 g/mol
That means that 1 mol of gallium weights 69.72 g and 1 mol of arsenic weights 74.92 g.
So, if 1 mol of gallium weights 69.72 g, the 4.00 g that we have will weight "x" g. Let's solve this using the "Rule of three":
69.72 g -------------- 1 mol of gallium
4.00 g -------------- x = (4.00 g . 1mol)/69.72 g = 0.057 moles of gallium
With the same reasoning we can find the moles of arsenic:
74.92 g -------------- 1 mol of arsenic
5.50 g -------------- x = (5.50 g . 1mol)/74.92 g = 0.073 moles of arsenic
To find the limiting and excess reactants it is very important to follow what the chemical reaction shows.
- If 1 mol of gallium needs 1 mol of arsenic to react, the 0.057 moles of gallium that we have will need 0.057 moles of arsenic, and we have 0.073 moles of aersenic, so we have more arsenic that what we need, that means that arsenic is the excess reactant.
- We can do the same analysis with arsenic: if 1 mol of arsenic reacts with 1 mol of gallium, the 0.073 moles of arsenic will react with 0.073 moles of gallium, but we only have 0.057 moles of gallium, so gallium is the limiting reactant.
3rd) Once you find the limiting and excess reactants, you have make the following calculus based on the limiting reactant, because is the one that if it runs out, the reaction will stop.
Up to here, we know that there are 0.057 moles of gallium and 0.073 moles of arsenic, and that 1 mol of gallium reacts with 1 mol of arsenic, so the 0.057 moles of gallium will react with 0.057 moles of arsenic, and at the end of the reaction there will be 0.073 moles - 0.057 moles = 0.016 moles of arsenic.
And at the end of the reaction there will be 0 g of gallium (because it is the limiting reactant and it is all consumed in the reaction).
4th) Finally, find the grams of arsenic from the number of moles that are left. If 1 mol of arsenic weights 74.92 g, the 0.016 moles of arsenic that are left will weight "x" g. Let's use the Rule of three here:
1 mol of arsenic ------------ 74.92 g
0.016 moles of arsenic ------------ x = 1.20 g
Are left 1.20 g of arsenic (excess reactant).