Tolong bantuannya no.12 - 15

[ 12 ] n CO2 = m / Mr = 1,76 gr / 44 = 0,04 mol
n (C2H5)2Zn yg bereaksi dgn O2 = (koef. (C2H5)2Zn / koef. CO2 ) × n CO2 = ¼ × 0,04 mol = 0,01 mol

n (C2H5)2Zn total = m / Mr = 6,17 gr / 123,4 = 0,05 mol

n (C2H5)2Zn yg bereaksi dgn H2O = 0,05 mol - 0,01 mol = 0,04 mol

n C2H6 = ( koef. C2H6 / koef. (C2H5)2Zn ) × n (C2H5)2Zn = 2/1 × 0,04 mol = 0,08 mol

V C2H6 = VM × n = 22,4 L/mol × 0,08 mol = 1,792 L ( B )


[ 13 ] n NO / V NO = n SO2 / V SO2
( 1 gr / 30 ) / 1,6 L = ( 4 gr / 64 ) / V = 3 L ( A )


[ 14 ] CxHy + O2 => CO2 + H2O
....... 100 mL ......... 400 mL 500 mL
100 : 400 : 500
1 : 4 : 5
x = 4
y = 2(5) = 10
CxHy = C4H10 => alkana (hidrokarbon jenuh) , memenuhi rumus C_n H_{2n+2}

Setarakan reaksinya :
C4H10 + 13/2 O2 => 4 CO2 + 5 H2O
100 mL ... 650 mL ... 400 mL .. 500 mL

Jawaban : D. (1), (2), (3), (4) BENAR


[ 15 ] x N2 + y O2 => 2 NxOy
.......... 10 L ... 15 L .......

10 : 15 = 2 : 3

x = 2
y = 3
NxOy = N2O3 ( D )