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3.0 g of magnesium was added to 12.0g of ethanoic

Berikut ini adalah pertanyaan dari murabetfiona pada mata pelajaran Kimia untuk jenjang Sekolah Dasar

3.0 g of magnesium was added to 12.0g of ethanoic acid. Mg + 2CH 3 COOH → (CH3COO)2Mg + H 2 The mass of one mole of Mg is 24g. The mass of one mole of CH3COOH is 60g.Which one, magnesium or ethanoic acid, is in excess? You must show you nasoning ​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

mass of Mg = 3 g

Atomic relative Mg (Ar) = 24 g/mole

mass of CH3COOH = 12 g

Molecular relative of CH3COOH (Mr) = 60 g/mole

Reaction occurs:

Mg + 2CH3COOH → (CH3COO)2Mg + H2 (balance)

  • first, calculate mole for each reactant

n \: = \frac{mass}{Ar/Mr}

n \: Mg= \frac{3 \: g}{24 \: g/mole} = 0.125 \: mole

n \: CH _{3}COOH = \frac{12 \: g}{60 \: g/mole} = 0.2 \: mole

  • identifying the limiting reactant from balanced equation, use the mole from calculation before and use coefficient for each reactant from balanced equation

limiting \: reactant = \frac{mole \: reactant}{coefficient \: of \: reactant}

limiting \: Mg = \frac{0.125}{1} = 0.125 \: mole

limiting \: CH_{3}COOH = \frac{0.2}{2} = 0.1 \: mole

use CH3COOH as limiting reactant because it has smallest mole quantity, so for 1 coefficient of each compoundcompound is 0,1 mole.

  • calculate the remaining reactant

Mg + 2CH3COOH → (CH3COO)2Mg + H2

initial 0.125 0.2 - -

reacted 0.1 0.2 0.1 0.1

excess 0.025 - 0.1 0.1

  • calculate mass of excess reactant, Mg

mass \: = excess \: mole \: \times Ar = 0.025 \: mole \times 24 \: g/mole = 0.6 \: g

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Last Update: Wed, 15 Jun 22
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