Respuesta :
Answer: 2.64 g of carbon dioxide that could be produced by the chemical reaction.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} CH_4=\frac{0.96g}{16g/mol}=0.06moles[/tex]
[tex]\text{Moles of} O_2=\frac{6.37}{32}=0.20moles[/tex]
[tex]CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)[/tex]
According to stoichiometry :
1 mole of [tex]CH_4[/tex] require = 2 moles of [tex]O_2[/tex]
Thus 0.06 moles of [tex]CH_4[/tex] will require=[tex]\frac{2}{1}\times 0.06=0.12moles[/tex] of [tex]O_2[/tex]
Thus [tex]CH_4[/tex] is the limiting reagent as it limits the formation of product and [tex]O_2[/tex] is the excess reagent.
As 1 moles of [tex]CH_4[/tex] give = 1 mole of [tex]CO_2[/tex]
Thus 0.06 moles of [tex]CH_4[/tex] give =[tex]\frac{1}{1}\times 0.06=0.06moles[/tex] of [tex]CO_2[/tex]
Mass of [tex]CO_2=moles\times {\text {Molar mass}}=0.06moles\times 44g/mol=2.64g[/tex]
Thus 2.64 g of carbon dioxide that could be produced by the chemical reaction.
