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
I don't know where the 42.094 came from.
8.00 mole NH3.
Convert to moles oxygen.
8 x (3 moles O2/4 moles NH3) = 6.00 moles oxygen.
Now convert 6.00 moles oxygen to grams. g = moles x molar mass.
96 gm.
To find the number of grams of O2 needed to react with 8.00 mol NH3, you can use the given balanced equation:
4NH3(g) + 3O2(g) -> 2N2(g) + 6H2O(g)
From the equation, we can see that the ratio between NH3 and O2 is 4:3. Therefore, for every 4 moles of NH3, we need 3 moles of O2.
Now, let's use this ratio to calculate the grams of O2 needed:
8.00 mol NH3 x (3 mol O2 / 4 mol NH3) x (32.00 g O2 / 1 mol O2) = 8.00 x 3 x 32.00 / 4 = 64.00 x 32.00 / 4 = 512.00 / 4 = 128.00 g O2
Therefore, 128.00 grams of O2 are needed to react with 8.00 mol of NH3.
To find the number of grams of O2 needed to react with 8.00 mol NH3, we can use the given balanced chemical equation and perform a conversion.
Step 1: Start with the given quantity in moles of NH3 (8.00 mol NH3).
Step 2: Use the molar ratio from the balanced equation to convert moles of NH3 to moles of O2. In this case, the ratio is 4 moles of NH3 to 3 moles of O2.
Step 3: Now, convert the moles of O2 to grams of O2 using the molar mass of O2, which is 32.00 g/mol.
Putting it all together, the calculation looks like this:
8.00 mol NH3 x (3 mol O2 / 4 mol NH3) x (32.00 g O2 / 1 mol O2) = 269 g O2
Therefore, 269 grams of O2 are needed to react with 8.00 moles of NH3.