If 7.00g of CH2 is burned in excess Oxygen, what number of moles is formed with H2O
mol of CH2 = 12+2 = 14 grams
so
7/14 = .5 mol of CH3
equation
2CH2 + 3O2 --> 2CO2 + 2H2O
so for every mole of CH2 we get one mole of water and one mole of CO2
so the result is .5 mole CO2 and .5 mole H2O
I wonder about CH2. What is it? It could be the empirical formula for C2H4 which is ethylene (now named ethene).
Perhaps intended CH4.
THANKS!!!!
To find the number of moles of H2O produced when 7.00g of CH2 is burned in excess oxygen, we need to use the balanced chemical equation and the molar masses of the compounds involved.
The balanced equation for the combustion of CH2 is:
CH2 + O2 → CO2 + H2O
From the equation, we can see that for every 1 mole of CH2, 1 mole of H2O is produced. Therefore, the number of moles of H2O formed will be the same as the number of moles of CH2 used.
To calculate the number of moles, follow these steps:
1. Find the molar mass of CH2:
- C has a molar mass of 12.01 g/mol, and H has a molar mass of 1.01 g/mol.
- So, the molar mass of CH2 is (12.01 g/mol x 1) + (1.01 g/mol x 2) = 14.03 g/mol.
2. Calculate the number of moles of CH2:
- Divide the given mass of CH2 (7.00g) by its molar mass:
7.00 g / 14.03 g/mol = 0.499 mol
Therefore, when 7.00g of CH2 is burned in excess oxygen, 0.499 moles of H2O are formed.