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[ 53 ] V = 1,2 L (STP)
% CH4 = 60% v/v
V CH4 = 60% × 1,2 L = 0,72 L

.. CH4 + 2 O2 => CO2 + 2 H2O
0,72 L .. 1,44 L .. 0,72 L ... 1,44 L

% C2H6 = 40% v/v
V C2H6 = 40% × 1,2 L =
0,48 L

C2H6 + 7/2 O2 => 2 CO2 + 3 H2O
0,48 L .. 1,68 L ....... 0,96 L ... 1,44 L

V CO2 = 0,72 L + 0,96 L = 1,68 L

Ca(OH)2 + CO2 => CaCO3 + H2O
...1,68 L .. 1,68 L ...... 1,68 L ... 1,68 L

n CaCO3 = V / 22,4 L/mol
= 1,68 L / 22,4 L/mol
= 0,075 mol

Mr CaCO3 = Ar Ca + Ar C + 3 Ar O
= 40 + 12 + 3(16)
= 100

m CaCO3 = Mr × n
= 100 × 0,075 mol
= 7,5 gram (A)

[ 54 ]
Asumsikan :
[NO] awal = x
[H2] awal = y
r awal = r

x => 2y => 2v
½ x => y => ¼ v

Persamaan laju reaksi :
r = k [NO]^a [H2]^b

a = orde NO
b = orde H2
orde reaksi = a + b

orde x :
( k [NO]^a [H2]^b ) / ( k [NO]^a [H2]^b ) = r1 / r2
( (½ x)^a × y^b ) / ( x^a × y^b ) = ¼ r / r
½^a = ¼
a = 2

orde y :
( k [NO]^a [H2]^b ) / ( k [NO]^a [H2]^b ) = v1 / v2
( x^a × (2y)^b ) / ( x^a × y^b ) = 2r / r
2^b = 2
b = 1

Persamaan laju reaksi :
r = k [NO]² [H2]

Jawaban : C