This is a limiting reagent (LR) problem. You know that because amounts are given for BOTH reactants.
C2H5OH + 3O2 ==> 2CO2 + 3H2O
g ethanol = 4.60 mL x 0.789 = approx 3.6 g BUT you need to this calculation and ALL that follow over to obtain a better answer for each step. Mine are just estimates.
mols ethanol = grams/molar mass = approx 0.08
How much H2O would that produce. That's
approx 0.8mols x (3 mols H2O/1 mol ethanol) = apprx 0.08 x 3 = approx 0.24 mols
Now for the O2.
mols O2 = grams/molar mass = approx 0.5
How much H2O would that produce? That's approx 0.5 x 3 mols H2O/3 mols O2) = approx 0.5 mols O2
You see we have two different values for mols H2O and only one can be right. The correct value in LR problems is ALWAYS the smaller value; therefore, we should produce approx 0.24 mol H2O from the ethanol and convert that to grams = approx 4.3 grams. (That's 0.24 mol x 18 g/mol = about 4.3 g). That is the theoretical yield (TY). The actual yield is 3.70 mL which with a density of 1.00 g/mL is 3.70 g and that's the actual yield (AY)
% yield = (AY/TY)*100 = ?
The combustion of liquid ethanol (C2H5OH) produces carbon dioxide and water. After 4.60mL of ethanol (density=0.789g/ml) was allowed to burn in the presence of 15.45g of oxygen gas, 3.70mL of water (density=1.00g/ml) was collected.
a)Determine the theoretical yield of H2O for the reaction.
b)Determine the percent yield of H2O for the reaction.
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