Answer:
3.32 g MgO
Explanation:
We have the masses of two reactants, so this is a limiting reactant problem.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. Gather all the information in one place with molar masses above the formulas and masses below them.
M_r: 24.30 32.00 40.30
2Mg + O₂ ⟶ 2MgO
Mass/g: 6.40 1.32
===============
Step 2. Calculate the moles of each reactant
Moles of Mg = 6.40 × 1/24.30
Moles of Mg= 0.2634 mol Mg
Moles of O₂ = 1.32 × 1/32.00
Moles of O₂ = 0.041 25 mol O₂
===============
Step 3. Identify the limiting reactant
Calculate the moles of MgO we can obtain from each reactant.
From Mg
:
The molar ratio is 2 mol MgO:2 mol Mg
Moles of MgO = 0.2634 × 2/2
Moles of MgO = 0.2634 mol MgO
From O₂:
The molar ratio is 2 mol MgO:1 mol O₂
Moles of MgO = 0.041 25 × 2/1
Moles of MgO = 0.082 50 mol MgO
The limiting reactant is O₂ because it gives the smaller amount of MgO.
===============
Step 4. Calculate the mass of MgO that you can obtain from O₂.
Mass of MgO = 0.082 50 × 40.30/1
Mass of MgO = 3.32 g MgO
