Percentage error = ((12.7/11.5)-100%)x 100% = 10.4% (3 s.f.)
Since the measured yield exceeds the theoretical one, this means the error is in excess. Divide the measured yield by the theoretical yield and subtract 100% to find the difference, then multiply by 100% to find the percentage.
The grams of Na that are needed to complete to react with 40..0 g of O2 is calculated as below
find the moles of O2 used = mass/molar mass
= 40 g/32g/mol = 1.25 moles
write the reacting equation
4Na+ O2 = 2Na2O
by use of mole ratio between Na to O2 which is 4 :1
the moles of Na = 1.25 x 4 = 5 moles
mass of Na = mass x molar mass
= 5 moles x 23 g /mol= 115 moles
The balanced reaction would be:<span>
C12H22O11 + 12O2 = 12CO2 + 11H2O
We are given the amount of oxygen used in the combustion. This will be the starting point of our calculation. We use the ideal gas equation to find for the number of moles.
n = PV / RT = 1.00(250 L) / (0.08206 atm L/mol K ) 273 K
n= 11.16 mol O2
</span>11.16 mol O2<span> (12 mol CO2 / 12 mol O2) = 11.16 mol CO2
V = nRT/P =</span>11.16 mol CO2<span> x 273 K x 0.08206 atm L/mol K / 1 atm
V=250 L</span>
