Question

Ammonia reacts with oxygen to form nitric acid oxide and water vapor. What is the theoretical yield of water vapor if you start with 20.0 g of ammonia and 50.0 g of oxygen?Identify the limiting reactant. 4NH3 + 5O2 —> 4NO + 6H2O

Answer 1

amount of water produced = 31.76 g of water

Step-by-step explanation:

20 g of ammonia has
`(20)/(17)` moles of ammonia

and 50 g of Oxygen has
`(50)/(32)` moles of oxygen .

to find the limiting agent we have to find the mole ratio

mole ratio of ammonia =
`(20)/(17* 4)=0.29`

mole ratio of oxygen =
`(50)/(32* 5)=0.31\\`

therefore the limiting reagent is ammonia.

4 moles ammonia gives 6 moles of water

therefore 1 mole of ammonia gives
`(6)/(4)` moles of water

therefore
`(20)/(17)` moles of ammonia gives
`(6)/(4)* (20)/(17)` moles of water.

amount of water produced = 1.76 moles of water = 31.76 g of water