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.
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
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.
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.
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.