Respuesta :
Answer:
A) Ethanol is the Limiting reactant
B) 5.36 grams
Explanations:
The chemical reaction for the chemical combustion of ethanol is expressed as:
[tex]C_2H_5OH(l)+3O_2(g)\rightarrow2CO_2(g)+3H_2O(l)[/tex]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
[tex]\begin{gathered} mole\text{ of ethanol}=\frac{mass}{molar\text{ mass}}=\frac{4.58}{46.07} \\ mole\text{ of ethanol}=0.0993moles \\ \end{gathered}[/tex]For the mole of oxygen
[tex]\begin{gathered} mole\text{ of oxygen}=\frac{12.5}{32} \\ mole\text{ of oxygen}=0.3906mole \\ for\text{ 1 atom: }\frac{0.3906}{3}=0.1302mole \end{gathered}[/tex]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:
[tex]\begin{gathered} mole\text{ of water}=3\times0.0993 \\ mole\text{ of water}=0.2979moles \\ Mass\text{ of water produced}=mole\times molar\text{ mass} \\ Mass\text{ of water produced}=0.2979\times18=5.36grams \end{gathered}[/tex]Therefore the theoretical yield of H2O is 5.36grams
