How many grams of O2 are needed to react with 18.2 g of NH3?

44 g oxygen are needed.

Explanation:

Given data:

Mass of oxygen needed = ?

Mass of ammonia = 18.2 g

Solution:

Chemical equation:

4NH₃ + 5O₂   →  4NO + 6H₂O

Now we will calculate the number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 18.2 g/ 17 g/mol

Number of moles = 1.1 mol

Now we will compare the moles of ammonia with oxygen from balance chemical equation.

                        NH₃              :                O₂

                          4                 :                 5

                          1.1                :              5/4×1.1 = 1.375 mol

Mass of oxygen needed:

Mass = number of moles × molar mass

Mass = 1.375 mol × 32 g/mol

Mass = 44 g

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied