If 225.O grams of NO react with excess NH3,how many grams of H2O are produced
The molecular formula is as followed:
6NO + 4NH3 --->5N2 + 6H2O
The ratio of moles of NO to H2O is 1 to 1.
225.0g of NO/30.01 g of NO/mol= moles of NO
Moles of NO=Moles of H20
Moles of H2O *18.02g of H2O/mol= mass of H2O
To find out how many grams of H2O are produced when 225.0 grams of NO reacts with excess NH3, we need to first balance the chemical equation of the reaction. The balanced equation for the reaction between NO and NH3 can be represented as:
4 NH3 + 6 NO -> 5 N2 + 6 H2O
From the balanced equation, we can see that 6 moles of H2O are produced for every 6 moles of NO consumed in the reaction. This means that the mole ratio between H2O and NO is 1:1.
To calculate the number of moles of NO, we divide the given mass of NO by its molar mass. The molar mass of NO is the sum of the molar masses of nitrogen (N) and oxygen (O), which is 14.01 g/mol + 16.00 g/mol = 30.01 g/mol.
Moles of NO = Mass of NO / Molar mass of NO
Moles of NO = 225.0 g / 30.01 g/mol = 7.499 moles
Since the mole ratio between H2O and NO is 1:1, we can conclude that 7.499 moles of H2O are produced.
To convert moles of H2O to grams, we multiply the number of moles by the molar mass of H2O. The molar mass of H2O is the sum of the molar masses of hydrogen (H) and oxygen (O), which is 1.008 g/mol + 16.00 g/mol = 17.01 g/mol.
Mass of H2O = Moles of H2O * Molar mass of H2O
Mass of H2O = 7.499 moles * 17.01 g/mol = 127.79 grams
Therefore, approximately 127.79 grams of H2O are produced when 225.0 grams of NO reacts with excess NH3.