What mass of oxygen is required to burn 28.8 g of ammonia?

Answer 1

The balanced equation is: 4NH^3 + 7O^2 -> 4NO^2 + 6H^2O

(please forgive the exponential values being superscript vs subscript...but you get point)

determine molar mass (M) of ammonia: M (NH^3) = 17.0 g/mol

determine how many moles of ammonia in 28.8g NH^3:

28.8g / 17.0 g-mol = 1.69 mol NH^3

remember: mol = M / m(g) (moles = molar mass divided by mass, in grams)

and the Molar ratio, which is: 7/4 (that is, you need 7 mol of O^2

to completely react with 4 mol of NH^3.

so, work out the numbers: 1.69 mol NH^3 * 7/4 = 2.9575 mol O^2

Once you have your number of moles of O^2 (2.9575), you multiply this by the Molar mass of O^2, which is 15.99 * 2 = ~32.0 g/mol

lastly, multiply the number of moles (mol) O^2, by Molar mass:

2.9575 mol * 32.0 g/mol = ~94.64 g O^2

Therefore, you need 94.64g O^2 to react completely with 28.8g NH^3.