Answer:
No ethane will be left after reaction
Explanation:
Balanced equation:[tex]2C_{2}H_{6}+7O_{2}\rightarrow 4CO_{2}+6H_{2}O[/tex]
According to balanced equation, 7 moles of [tex]O_{2}[/tex] react completely with 2 moles of [tex]C_{2}H_{6}[/tex]
Molar mass of [tex]O_{2}[/tex] = 32 g/mol and molar mass of ethane = 30.07 g/mol
So, 48 g of [tex]O_{2}[/tex] = [tex]\frac{48}{32}mol[/tex] of [tex]O_{2}[/tex] = 1.5 moles of [tex]O_{2}[/tex]
Hence, 1.5 moles of [tex]O_{2}[/tex] react completely with [tex](\frac{2}{7}\times 1.5)[/tex] moles of [tex]C_{2}H_{6}[/tex] or 0.43 moles of [tex]C_{2}H_{6}[/tex]
So, mass of [tex]C_{2}H_{6}[/tex] reacts completely with 1.5 moles of [tex]O_{2}[/tex] = [tex](0.43\times 30.07)g[/tex] = 12.9 g
As only 8.42 g of ethane has been placed in reaction mixture therefore no ethane will be left after reaction.
