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Ruas Kanan

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Kimia X SMA 89<br />

Jawab:<br />

C H (g) +5 O(g) 3 8 2<br />

⎯⎯→ 3 CO (g) + 4 H O(l)<br />

2 2<br />

a.<br />

Volume O2<br />

Volume C3H 8<br />

=<br />

koefisien O2<br />

koefisien C3H8 Volume O2 =<br />

koefisien O2<br />

koefisien C3H 8<br />

× volume C3H8 = 5<br />

× 3 liter = 15 liter<br />

1<br />

b.<br />

Volume CO2<br />

Volume C3H 8<br />

=<br />

koefisien CO2<br />

koefisien C3H8 Volume CO2 =<br />

koefisien CO2<br />

koefisien C3H 8<br />

× volume C3H8 = 3<br />

× 3 liter = 9 liter<br />

1<br />

c.<br />

Volume H2O Volume C3H 8<br />

=<br />

koefisien H2O koefisien C3H8 Volume H O 2 =<br />

koefisien H2O koefisien C3H 8<br />

× volume C3H8 = 4<br />

× 3 liter = 12 liter<br />

1<br />

2. Sepuluh mL gas nitrogen (N ) dan 15 mL gas oksigen (O ) tepat habis bereaksi menjadi<br />

2 2<br />

10 mL gas N O . Tentukan rumus kimia gas N O tersebut!<br />

a b a b<br />

Jawab:<br />

Perbandingan koefisien = perbandingan volume<br />

Koefisien N : O : N O = 10 : 15 : 10 = 2 : 3 : 2<br />

2 2 a b<br />

2 N 2 + 3 O 2 ⎯⎯→ 2 N a O b<br />

Karena jumlah atom di ruas kiri dan di ruas kanan sama, maka harga a dan b dapat<br />

dicari sebagai berikut.<br />

Jumlah atom N kiri= Jumlah atom N kanan<br />

2 × 2 = 2a<br />

4 = 2a<br />

a = 2<br />

Jumlah atom O kiri= Jumlah atom O kanan<br />

3 × 2 = 2b<br />

6 = 2b<br />

b = 3<br />

Jadi, rumus kimia senyawa tersebut adalah N 2 O 3 .

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