Ethanol, C2H5OH burns with oxygen in air to give carbon dioxide and water. what is the amount (in moles) of water produced from 0.51 mol C2H5OH.the equation is C2H5OH + 3O2 -----> 2CO2 + 3H2O
This is an example of a combustion reaction. To answer this you must first balance the equation.
The easiest way is to set up the reaction equation without any coefficients:
C2H6O + O2 --> CO2 + HOH
Now start balancing, first with carbon, then with hydrogen and last the oxygen.
There are 2 carbons on the left and 6 hydrogens so you add a coefficient of 2 in front of the CO2 and a 3 in front of the water.
C2H6O + O2 --> 2 CO2 + 3 HOH
Now count the number of oxygen on the product side and you get seven. One oxygen came from the ethanol so you need 6 more from oxygen. Add a coefficient of 3 in front of the oxygen on the reactant side and your reaction is balanced.
C2H60 + 3 O2 --> 2 CO2 + 3 HOH
This shows you that for every mole of ethanol, you will get 3 moles of water. So if you start with 0.51 moles of ethanol, you will produce 3 x .51 or 1.53 moles of water.
The combustion equation is C2H5OH +3O2 = 2CO2 + 3H2O.
The given equation is a balanced equation.
To find the amount of mole of water produced when 0.51 mole of Ethanol , C2H5OH is burnt and cobines with oxygen.
The equation self explains that :
One mole of (C2H5OH) +3mole of (O2) results in products of 2 mole of (CO2) and 3moles of(H2O).....(1)
Threfore if you multiply the equation at (1) by 0.51 we get:
0.51 mole of (C2H%OH) + 3*(0.51)mole of (O2(O2) results in products of 2*(0.51) mole of (CO2) and 3*(0.51) mole of (H2O).
So the equation answers that 3*0.51 mole of H2O is produced.
3*0.51 mole of H2O = 1.53 mole of water is produced when 0.51 mole of Ethanol is combined(burned in) with oxygen.
x=1.53mole of H2O
0.51mole C2H5OH/ 1mol=mole H20/ 3mol
x= 1.53 mole of H2O
1.53 moL of H20
C2H5OH + 3O2 -----> 2CO2 + 3H2O
amount (in moles) of water produced from 0.51 mol C2H5OH?
XmoLH2O / 3moL = 0.51moLC2H5OH / 1moL
- now we cross multiply
1.53 moL of water is produced from 0.51 mol C2H5OH.
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