Ammonia and oxygen react to form nitrogen and water.
4NH3(g)+3 O2(g)-2N2(g)+6H2O(g)
a.How many grams of O2 are needed to react with 8.00 mol NH3?
8.00 mol NH3 x42.094/4 mol NH3 x32.00 O2/1 mol O2=269gO2
Ammonia and oxygen react to form nitrogen and water.
4NH3(g)+3 O2(g)-2N2(g)+6H2O(g)
a.How many grams of O2 are needed to react with 8.00 mol NH3?
8.00 mol NH3 x42.094/4 mol NH3 x32.00 O2/1 mol O2=269gO2
To determine how many grams of O2 are needed to react with 8.00 mol NH3, you can use the stoichiometry of the balanced equation.
From the balanced equation: 4 NH3(g) + 3 O2(g) -> 2 N2(g) + 6 H2O(g)
We can see that the molar ratio between NH3 and O2 is 4:3. Therefore, for every 4 moles of NH3, we need 3 moles of O2.
First, convert the given 8.00 mol NH3 to moles of O2:
8.00 mol NH3 x (3 mol O2 / 4 mol NH3) = 6.00 mol O2
Now, we need to convert the moles of O2 to grams. The molar mass of O2 is 32.00 g/mol.
6.00 mol O2 x 32.00 g O2 / 1 mol O2 = 192 g O2
So, 8.00 mol NH3 reacts with 192 grams of O2.
To calculate the grams of O2 needed to react with 8.00 mol of NH3, we can use the given balanced equation and stoichiometry.
First, we need to find the molar ratio between NH3 and O2. According to the equation, 4 moles of NH3 react with 3 moles of O2.
So, 4 mol NH3 : 3 mol O2.
Next, we can set up a conversion factor to convert from moles of NH3 to moles of O2:
8.00 mol NH3 x (3 mol O2 / 4 mol NH3) = 6.00 mol O2.
Now, we can convert the moles of O2 to grams using the molar mass of O2:
6.00 mol O2 x (32.00 g O2 / 1 mol O2) = 192 g O2.
Therefore, 8.00 mol of NH3 requires 192 grams of O2 to react.