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Question
how does the law of conservation of mass apply to this reaction: $c_{2}h_{4}+3o_{2}\to2h_{2}o + 2co_{2}$?
each element needs to be balanced.
the equation needs to be balanced. there are fewer oxygen atoms in the equation than hydrogen or carbon.
only the oxygen needs to be balanced. there are equal numbers of hydrogen and carbon.
the law of conservation of mass has already been applied. there is an equal number of each element on both sides of the equation.
According to the law of conservation of mass, in a chemical reaction, the mass of reactants must equal the mass of products. This means the number of atoms of each element on the reactant side must be equal to the number of atoms of each element on the product side. Let's check each element in the given reaction \(C_{2}H_{4}+3O_{2}\to2H_{2}O + 2CO_{2}\):
- Carbon (\(C\)):
- Reactant side: In \(C_{2}H_{4}\), the number of \(C\) atoms is \(2\).
- Product side: In \(2CO_{2}\), the number of \(C\) atoms is \(2\times1 = 2\).
- Hydrogen (\(H\)):
- Reactant side: In \(C_{2}H_{4}\), the number of \(H\) atoms is \(4\).
- Product side: In \(2H_{2}O\), the number of \(H\) atoms is \(2\times2=4\).
- Oxygen (\(O\)):
- Reactant side: In \(3O_{2}\), the number of \(O\) atoms is \(3\times2 = 6\).
- Product side: In \(2H_{2}O+2CO_{2}\), the number of \(O\) atoms is \((2\times1)+(2\times2)=2 + 4=6\).
Since the number of atoms of \(C\), \(H\), and \(O\) is the same on both sides of the equation, the law of conservation of mass has already been applied.
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The law of conservation of mass has already been applied. There is an equal number of each element on both sides of the equation.