Answer:
3.32 g MgO
Explanation:
We have the masses of two reactants, so this is a <em>limiting reactant problem</em>.
We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.
Step 1. <em>Gather all the information in one place</em> 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
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Step 2. Calculate the <em>moles of each reactant </em>
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
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Step 3. I<em>dentify the limiting reactant </em>
Calculate the moles of MgO we can obtain from each reactant.
<em>From Mg</em>
:
The molar ratio is 2 mol MgO:2 mol Mg
Moles of MgO = 0.2634 × 2/2
Moles of MgO = 0.2634 mol MgO
<em>From O₂</em>:
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 <em>limiting reactant is O₂</em> because it gives the smaller amount of MgO.
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Step 4. Calculate the <em>mass of MgO</em> that you can obtain from O₂.
Mass of MgO = 0.082 50 × 40.30/1
Mass of MgO = 3.32 g MgO