Pure magnesium metal is often found as ribbons and can easily burn in the presen

Question

Pure magnesium metal is often found as ribbons and can easily burn in the presence of oxygen. When 4.87 g of magnesium ribbon burns with 8.57 g of oxygen, a bright, white light and a white, powdery product are formed. Enter the balanced chemical equation for this reaction. Be sure to include all physical states.

1. Write a balanced equation.

2, What is the limiting reactant?

3. The percent yield for the reaction is 81.2%, how many grams of product were recovered?

4. How many grams of excess reactant remain?

Explanation / Answer

1)
Balanced chemical equation is:
2 Mg + O2 ---> 2 MgO

2)
Molar mass of Mg = 24.31 g/mol


mass(Mg)= 4.87 g

use:
number of mol of Mg,
n = mass of Mg/molar mass of Mg
=(4.87 g)/(24.31 g/mol)
= 0.2003 mol

Molar mass of O2 = 32 g/mol


mass(O2)= 8.57 g

use:
number of mol of O2,
n = mass of O2/molar mass of O2
=(8.57 g)/(32 g/mol)
= 0.2678 mol

Balanced chemical equation is:
2 Mg + O2 ---> 2 MgO

2 mol of Mg reacts with 1 mol of O2
for 0.2003 mol of Mg, 0.1002 mol of O2 is required
But we have 0.2678 mol of O2

so, Mg is limiting reagent

3)
we will use Mg in further calculation


Molar mass of MgO,
MM = 1*MM(Mg) + 1*MM(O)
= 1*24.31 + 1*16.0
= 40.31 g/mol

According to balanced equation
mol of MgO formed = (2/2)* moles of Mg
= (2/2)*0.2003
= 0.2003 mol


use:
mass of MgO = number of mol * molar mass
= 0.2003*40.31
= 8.075 g

% yield = actual mass*100/theoretical mass
81.2= actual mass*100/8.075
actual mass=6.56 g

Answer: 6.56 g

4)
According to balanced equation
mol of O2 reacted = (1/2)* moles of Mg
= (1/2)*0.2003
= 0.1002 mol
mol of O2 remaining = mol initially present - mol reacted
mol of O2 remaining = 0.2678 - 0.1002
mol of O2 remaining = 0.1676 mol


Molar mass of O2 = 32 g/mol

use:
mass of O2,
m = number of mol * molar mass
= 0.1676 mol * 32 g/mol
= 5.36 g
Answer: 5.36 g

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