Ответ:

дано

m(CH3Br) = 3.85 g

m(Na) = 1.15 g

V пр (C2H6) = 385.4 ml = 0.3854 L

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η(C2H6)-?

2CH3Br+2Na-->C2H6+2NaBr

M(CH3Br) = 95 g/mol

n(CH3Br) = m/M = 3.85 / 95 = 0.04 mol

M(Na)= 23 g/mol

n(Na)= m/M = 1.15 / 23 = 0.05 mol

n(CH3Br) < n(Na)

2n(CH3Br) = n(C2H6)

n(C2H6) = 0.04 / 2 = 0.02 mol

Vтеор(C2H6) = n(C2H6) * Vm = 0.02 * 22.4 = 0.448 L

η(C2H6) = V пр(C2H6) / V теор(C2H6) * 100% = 0.3584 / 0.448 * 100% = 80%

ответ 80%

Объяснение: