Answer:
0 g.
Explanation:
Hello,
In this case, since the reaction between methane and oxygen is:
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:
[tex]n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4[/tex]
Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.
[tex]left\ over=0g[/tex]
Regards.
