Respuesta :
1) Chemical reaction
CH3 CH2 OH + O2 ---> CH3COOH + H2O
It is balanced
2) Molar ratios
1 mol CH3 CH2OH : 1 mol O2 : 1 mol CH3COOH : 1 mol H2O
3) Conversion of 7.6 grams of ethanol to moles
Molar mass of CH3CH2OH = 2 * 12g/mol + 6 * 1 g/mol + 16 g/mol = 46 g/mol
moles = mass / molar mass = 7.6 g / 46g/mol = 0.165 mol
4) Use of proportions with the molar ratios:
1mol CH3CH2OH / 1mol O2 = 0.165 mol CH3CH2OH / x => x = 0.165 mol CH3CH2OH.
5) Conversion of 0.165 mol O2 to grams
mass in grams = number of moles * molar mass = 0.165 mol * (2 * 16g/mol)
mass = 5.28 g ≈ 5.3 g
Answer: 5.3 g
CH3 CH2 OH + O2 ---> CH3COOH + H2O
It is balanced
2) Molar ratios
1 mol CH3 CH2OH : 1 mol O2 : 1 mol CH3COOH : 1 mol H2O
3) Conversion of 7.6 grams of ethanol to moles
Molar mass of CH3CH2OH = 2 * 12g/mol + 6 * 1 g/mol + 16 g/mol = 46 g/mol
moles = mass / molar mass = 7.6 g / 46g/mol = 0.165 mol
4) Use of proportions with the molar ratios:
1mol CH3CH2OH / 1mol O2 = 0.165 mol CH3CH2OH / x => x = 0.165 mol CH3CH2OH.
5) Conversion of 0.165 mol O2 to grams
mass in grams = number of moles * molar mass = 0.165 mol * (2 * 16g/mol)
mass = 5.28 g ≈ 5.3 g
Answer: 5.3 g
The mass of oxygen required to react with 7.6 g of ethanol is 5.3 g.
Further Explanation
This problem is a stoichiometric problem. To solve problems like this we always begin with the balanced equation. Then, we use the stoichiometric ratios provided by the coefficients of the reactants and products.
1. Write the balanced chemical equation.
CH₃CH₂OH + O₂ → H₂O + CH₃COOH
2. Convert the mass of oxygen to moles.
[tex]7.6 \ g \ CH_3CH_2OH \times \frac{1 \ mol CH_3CH_2OH}{46.07 \ g} = 0.165 \ mol \ CH_3CH_2OH\\[/tex]
3. Determine the equivalent moles of oxygen that reacts with the given quantity of ethanol. The stoichiometric ratio for ethanol and oxygen indicated in the balanced chemical equation is 1:1.
[tex]moles \ O_2 = 0.165 \ mol \ CH_3CH_2OH \times \frac{1 \ mol O_2}{ 1 \ mol CH_3CH_2OH}\\\\moles \ O_2 = 0.165 \ mol \ O_2[/tex]
4. Convert the moles of O₂ to mass.
[tex]mass \ of \ O_2 = 0.165 \ mol \ O_2 \times \frac{32 \ g}{1 \ mol \ O_2 }\\\\\boxed {mass \ of \ O_2 \ = 5.28 \ g}[/tex]
The answer required should only have 2 significant digits. Therefore, the final answer is:
[tex]\boxed {\boxed {mass \ of \ O_2 \ = 5.3 \ g}}[/tex]
Learn More
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