Question

Given the following equation: 4 NH3 (g)5 O2 (g) >4 NO (g) + 6 H20 (I) How many moles of NH3 is required to react with 25.7 grams of O2?

Answer 1

0.6425 moles of
`NH_3` is required to react with 25.7 grams of
`O_2`.

Step-by-step explanation:

mas of oxygen gas = 25.7 g

moles of oxygengas =
`(25.7 g)/(32 g/mol)=0.8031 mol`

`4NH_3 (g)+5O_2 (g) \rightarrow 4 NO(g) + 6H_20 (I)`

According to reaction given above, 5 moles of oxygen gas reacts with 4 moles of ammonia gas.

Then 0.8031 moles of oxygen gas will react with :

`(4)/(5)* 0.8031 mol=0.6425 mol` of ammonia gas

0.6425 moles of
`NH_3` is required to react with 25.7 grams of
`O_2`.