Respuesta :
The percent yield of [tex]MgO[/tex] is 58.79%
According to the following balanced reaction, magnesium interacts with oxygen to generate magnesium oxide:
[tex]2Mg + O_2 \rightarrow 2MgO[/tex]
Given mass of solid magnesium = 0.261g
Molar mass of Mg = 24g/mol
Using formula , [tex]Number of moles =\dfrac {mass} {molar mass} = \dfrac {0.261g} {24g/mol} = 0.0108mole[/tex]
We could theoretically make 0.0108 moles of [tex]MgO[/tex] since the reaction yields an equimolar quantity of [tex]MgO[/tex]. It is clear from the above reaction that 2 moles of magnesium result in 2 moles of magnesium oxide.
Hence, mass of [tex]MgO[/tex] = Number of moles [tex]\times[/tex] Molar mass of MgO
mass of [tex]MgO[/tex] = [tex]0.0108mole \times 40g/mole = 0.432g[/tex]
Percent yield = [tex]\dfrac{Actual mass of MgO} {Theoretical mass of MgO} \times 100[/tex]
= [tex]\dfrac{0.254} {0.432} \times 100 = 58.79percentage[/tex]
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