Answer: The theoretical yield of magnesium oxide is 6.65 grams.
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of magnesium = 4.00 g
Molar mass of magnesium = 24.3 g/mol
Putting values in equation 1, we get:
[tex]\text{Moles of magnesium}=\frac{4.00g}{24.3g/mol}=0.165mol[/tex]
The given chemical equation follows:
[tex]2Mg(s)+O_2(g)\rightarrow 2MgO(s)[/tex]
By Stoichiometry of the reaction:
2 moles of magnesium produces 2 moles of magnesium oxide
So, 0.165 moles of magnesium will produce = [tex]\frac{2}{2}\times 0.165=0.165mol[/tex] of magnesium oxide
Now, calculating the theoretical yield of magnesium oxide by using equation 1:
Molar mass of magnesium oxide = 40.3 g/mol
Moles of magnesium oxide = 0.165 moles
Putting values in equation 1, we get:
[tex]0.165mol=\frac{\text{Mass of magnesium oxide}}{40.3g/mol}\\\\\text{Mass of magnesium oxide}=(0.165mol\times 40.3g/mol)=6.65g[/tex]
Hence, the theoretical yield of magnesium oxide is 6.65 grams.
The theroretical yield is 6.8 g.
The equation of the reaction is;
2Mg(s) + O2(g) ------>2MgO(s)
Number of moles of Mg reacted = 4 g/24 g/mol = 0.17 moles
Now;
2 moles of Mg yields 2 moles of MgO
0.17 moles of Mg yields 2 moles × 0.17 moles/2 moles = 0.17 moles of MgO
Mass of MgO produced = 0.17 moles × 40 g/mol = 6.8 g
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