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Question
essay: in the presence of sulfuric acid, one molecule of ethanol (c₂h₆o) can react to form one molecule each of water and another product: c₂h₆o ⟶ h₂o + ? using the law of conservation of mass, describe a technique a chemist could use to predict the formula of the unknown product; then use the technique you described to predict the formula of the unknown product.
Step1: Recall Law of Conservation of Mass
The law of conservation of mass states that in a chemical reaction, the total number of atoms of each element in the reactants must equal the total number of atoms of each element in the products. So we need to count the number of each type of atom in the reactant (\(C_2H_6O\)) and the known product (\(H_2O\)) and then find the unknown product that balances the atoms.
Step2: Count Atoms in Reactant
For the reactant \(C_2H_6O\):
- Number of carbon (\(C\)) atoms: \(2\)
- Number of hydrogen (\(H\)) atoms: \(6\)
- Number of oxygen (\(O\)) atoms: \(1\)
Step3: Count Atoms in Known Product
For the product \(H_2O\):
- Number of carbon (\(C\)) atoms: \(0\)
- Number of hydrogen (\(H\)) atoms: \(2\)
- Number of oxygen (\(O\)) atoms: \(1\)
Step4: Determine Atoms in Unknown Product
Let the unknown product be \(X\). Let the formula of \(X\) be \(C_xH_yO_z\).
For carbon: The number of \(C\) atoms in reactant is \(2\), in \(H_2O\) is \(0\), so in \(X\), \(x = 2\) (to balance \(C\) atoms: \(2=0 + x\)).
For hydrogen: The number of \(H\) atoms in reactant is \(6\), in \(H_2O\) is \(2\), so in \(X\), \(y=6 - 2=4\) (since \(6=2 + y\)).
For oxygen: The number of \(O\) atoms in reactant is \(1\), in \(H_2O\) is \(1\), so in \(X\), \(z = 1- 1=0\) (since \(1=1 + z\)).
So the unknown product has \(2\) \(C\) atoms, \(4\) \(H\) atoms, and \(0\) \(O\) atoms. The formula is \(C_2H_4\) (ethene). The technique is to balance the chemical equation by equating the number of each type of atom on both sides of the reaction arrow.
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The unknown product has the formula \(C_2H_4\) (ethene). The technique is using the law of conservation of mass to balance the number of each element's atoms between reactants and products: count atoms in reactant (\(C_2H_6O\): 2 C, 6 H, 1 O) and known product (\(H_2O\): 0 C, 2 H, 1 O), then find the unknown product (\(C_2H_4\)) that provides the remaining 2 C and 4 H atoms to balance the equation \(C_2H_6O
ightarrow H_2O + C_2H_4\).