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.

Answer 2

The balanced chemical equation for the combustion of ammonia is 4NH3 + 3O2 -> 2N2 + 6H2O. From the equation, you can see that 3 moles of O2 are required to burn 4 moles of NH3. First, convert the mass of NH3 to moles using its molar mass. Then, use the mole ratio from the balanced equation to find the moles of O2 needed, and finally convert that to grams.