1) Chemical equation
C2H5OH + 3 O2 ---> 2 CO2 + 3 H2O
2) Molar ratios
1 mol C2H5OH : 3 mol O2 : 2 mol CO2 : 3 mol H2O
3) Amount of ethanol burned
D = M / V => M = D * V = 0.789 g/ml * 4.61 ml = 3.637 g
Number of moles = mass in grams / molar mass
molar mass of C2H5OH = 2 *12 g/mol + 6*1.0 g/mol + 16.0 g/mol = 46.0 g/mol
=> number of moles of C2H5OH = 3.637 g / 46.0 g/mol = 0.079 mol
4) Amount of oxygen, O2
number of moles of O2 = mass in grams / molar mass
number of moles O2 = 15.70g / 32.0 g/mol = 0.49 mol
5) Limiting reactant
Theoretical ratio: 1 mol C2H5OH / 3 mol O2
Actual ratio: 0.079 mol / 0.49 mol = 0.16
=> there is more oxygen than needed to burn all the ethanol => ethanol burns completely and it is the limiting reactant.
6) Theoretical yield of H2O
1 mol C2H5OH / 3 mol H2O = 0.079 mol C2H5OH / x
=> x = 0.079 mol C2H5OH * 3 mol H2O / 1 mol C2H5OH
=> x = 0.237 mol H2O
Conversion to grams:
mass = molar mass * number of moles
mass H2O = 18.0 g/mol * 0.237 g/mol = 4.27 g <---- theoretical yield
7) grams of H2O collected (yield)
M = V * D = 3.27 ml * 1.00 g/ml = 3.27 g <----- actual yield
8) Percent yield
Percent yield = (actual yield / theoretical yield) * 100 = (3.27 g / 4.27) * 100 = 76.6%
Answer: 76.6%
Answer:
0.3811 mol.
Explanation:
- To calculate the no. of moles of a gas, we can use the general law of ideal gas: <em>PV = nRT</em>.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
∵ P = 2.1 atm, V = 4.5 L, T = 302 K, R = 0.0821 L.atm/mol.K.
<em>∴ n = PV/RT </em>= (2.1 atm)(4.5 L)/(0.0821 L.atm/mol.K)(302.0 K) = <em>0.3811 mol.</em>
Transportation seems like the right answer
Inches because that is a trait of measurement
