Question

The combustion of liquid ethanol (C2H, OH)produces carbon dioxide and water. After 5.8 mLof ethanol (density = 0.789 g/mL) was allowed toburn in the presence of 12.5 g of oxygen gas, 3.10mL of water (density = 1.00 g/mL) was collected.Part A: Determine the limiting reactantPart B. Determine the theoretical yield of H2O

Answer 1

A) Ethanol is the Limiting reactant

B) 5.36 grams

Explanations:

The chemical reaction for the chemical combustion of ethanol is expressed as:

`C_2H_5OH(l)+3O_2(g)\rightarrow2CO_2(g)+3H_2O(l)`

Determine the mass of ethanol

Mass of ethanol = density * volume

Mass of ethanol =0.789 * 5.8

Mass of ethanol = 4.58 grams

Determine the mass of Oxygen

Mass of Oxygen = 12.5 grams

Determine the mole of ethanol and oxygen

`\begin{gathered} mole\text{ of ethanol}=\frac{mass}{molar\text{ mass}}=(4.58)/(46.07) \\ mole\text{ of ethanol}=0.0993moles \\ \end{gathered}`

For the mole of oxygen

`\begin{gathered} mole\text{ of oxygen}=(12.5)/(32) \\ mole\text{ of oxygen}=0.3906mole \\ for\text{ 1 atom: }(0.3906)/(3)=0.1302mole \end{gathered}`

Since the mole of ethanol is lower than that of oxygen, hence the limiting reactant is ethanol

Part B: According to stoichiometry, 1mole of ethanol produces 3 moles of water, the mole of water required will be expressed as:

`\begin{gathered} mole\text{ of water}=3*0.0993 \\ mole\text{ of water}=0.2979moles \\ Mass\text{ of water produced}=mole* molar\text{ mass} \\ Mass\text{ of water produced}=0.2979*18=5.36grams \end{gathered}`

Therefore the theoretical yield of H2O is 5.36grams