Answer:
[tex]Excess=3.53g[/tex]
Explanation:
Hello,
In this case, the undergoing chemical reaction is:
[tex]2Mg(s)+O_2(g) \rightarrow 2MgO(s)[/tex]
Next, we identify the limiting reactant by computing the moles of magnesium oxide yielded by 3.86 g of magnesium and 155 mL of oxygen at the given conditions via their 2:1:2 mole ratios and the ideal gas equation:
[tex]n_{MnO}^{by \ Mg}=3.86gMg*\frac{1molMg}{24.3gMg}*\frac{2molMgO}{2molMg} =0.159molMgO\\\\n_{MnO}^{by \ O_2}=\frac{1atm*0.155L}{0.082\frac{atm*L}{molO_2*K}*275K} *\frac{2mol MgO}{1molO_2} =0.0137molMgO[/tex]
It means that the limiting reactant is the oxygen as it yields the smallest amount of magnesium oxide. Next, we compute the mass of magnesium consumed the oxygen only:
[tex]m_{Mg}^{consumed}=0.0137molMgO*\frac{2molMg}{2molMgO} *\frac{24.3gMg}{1molMg} =0.334gMg[/tex]
Thus, the mass in excess is:
[tex]Excess=3.86g-0.334g\\\\Excess=3.53g[/tex]
Regards!
