2.0 gram of C2H4 is reacted withO2. What mass of water is produced?
Hmm, well try to write a balanced equation
C2 H4 + 3O2 ---> 2CO2 + 2H2O
seems ok
so for every mol of C2H4 I get 2 mol water
C2H4 = 2*12 + 4*1 = 28 grams/mol
so
2 grams is 2/28 = 1/14 of a mol
so
I get 1/7 mol of water
water is 16+2 = 18 grams/mol
18/7 = 2.57 grams of water
2.57 grams of water
Hmm, well try to write a balanced equation
C2 H4 + 3O2 ---> 2CO2 + 2H2O
seems ok
so for every mol of C2H4 I get 2 mol water
C2H4 = 2*12 + 4*1 = 28 grams/mol
so
2 grams is 2/28 = 1/14 of a mol
so
I get 1/7 mol of water
water is 16+2 = 18 grams/mol
18/7 = 2.57 grams of water
To determine the mass of water produced, we need to first calculate the balanced equation for the reaction between C2H4 (ethylene) and O2 (oxygen).
The balanced equation for the complete combustion of ethylene can be written as:
C2H4 + 3O2 -> 2CO2 + 2H2O
From the balanced equation, we can see that for every 2 moles of C2H4, 2 moles of H2O are produced. Therefore, we need to calculate how many moles of C2H4 are present in 2.0 grams.
To calculate the number of moles, we can use the molar mass of C2H4, which is:
2 * (12.01 g/mol) + 4 * (1.01 g/mol) = 28.05 g/mol
Now, we can calculate the number of moles of C2H4 using the formula:
moles = mass / molar mass
moles of C2H4 = 2.0 g / 28.05 g/mol = 0.071 moles
From the balanced equation, we know that for every 2 moles of C2H4, 2 moles of H2O are produced. Therefore, if we have 0.071 moles of C2H4, we will also have 0.071 moles of H2O.
Finally, we can calculate the mass of water using the moles of water and its molar mass. The molar mass of water (H2O) is:
2 * (1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
mass of H2O = moles of H2O * molar mass of H2O
mass of H2O = 0.071 moles * 18.02 g/mol ≈ 1.28 grams
Therefore, approximately 1.28 grams of water will be produced when 2.0 grams of C2H4 reacts with O2.