Question

How do I do this limiting reactants problem about magnesium oxide?

Object
Mass
Weigh Boat
0.24 grams
Mass of MgO + Weigh Boat
3.79 grams
2nd weighing: Mass of MgO + Weigh Boat
3.5 grams

1.) What mass of MgO did the students actually collect? How did you find this?

3.55 grams, 3.79-0.24

2.) The students went in the lab with 2.40 grams of Mg and 5.52 grams of O2. Which of these reactants is the limiting reactant?

2.40g Mg(
5.52g O2(

3.) How much MgO should the students have found in the lab?

Answer 1

See below

Explanation:

Number 1 is not correct, always use the second weighing mass the sample of MgO had to be cooled before weighing, I suggest reweighing it again and again until the mass reads the same.

Addressing number 2:
Write a balanced equation
`2Mg+O_2->2MgO`

2)
Determine the limiting reagent:
`2.40 cancel("g Mg")* (1 cancel("mol Mg"))/ (24.3 cancel("g Mg")) *(1 cancel("mol O"_2))/ (2 cancel("mol Mg"))* (32 "g O"_2)/ (1 cancel("mol O"_2))= 1.58 g O_2` required to react, since we have that amount, `O_2` is in excess and Mg is said to be limiting

3)
Stoichometry using limiting reagent:
`2.40 cancel("g Mg")* (1 cancel("mol Mg"))/ (24.3 cancel("g Mg"))* (2 cancel("mol MgO"))/ (2 cancel("mol Mg"))* (40.3"g MgO")/ (1 cancel("mol MgO"))= 3.98 g MgO`

4) Percent Error
`(%"Error") = ("Actual - Theoretical")/("Theoretical") xx 100%`

`(%"Error") = |(3.26 - 3.98)/(3.98)| xx 100% = 18.1%`

Percent yield (optional):
`("Experimental yield")/("Theoretical yield")*100`
`(3.26 g)/ (3.98 g)*100= 81.9%`

Note: Your experimental yield should never exceed your theoretical yield or there is problem with your lab methodology