Calculate the mass of water produced when 42.0 g of propane, c3h8, is burned with 115 g of oxygen. hint: write balanced equation first

            51.75 g of H₂O

Solution:

              The balance chemical equation is as follow,

                                C₃H₈  +  5 O₂    →    3 CO₂  +  4 H₂O

Step 1: Finding Limiting Reagent:

According to equation,

               44.1 g (1 mol) C₃H₈ reacts with  =  160 g (5 mol) of O₂

So,

                        42 g C₃H₈ will react with  =  X g of O₂

Solving for X,

                     X  =  (42 g × 160 g) ÷ 44.1 g

                     X  =  152.38 g of O₂

As we are only provided with 115 g of O₂, Hence it is the limiting reactant and it will control the yield of products.

Step 2: Calculating amount of H₂O:

According to equation,

               160 g (5 mol) O₂ produces  =  72 g (4 mol) of H₂O

So,

                        115 g O₂ will produce  =  X g of H₂O

Solving for X,

                     X  =  (72 g × 115 g) ÷ 160 g

                     X  =  51.75 g of H₂O

a 20-cm-long, cylindrical graphite (pure carbon) rod is inserted into an oxidizing atmosphere at 1145 k and 1.013 x 105 pa pressure. the oxidizing process is limited by the diffu- sion of oxygen counterflow to the carbon monoxide that is formed on the cylindrical surface.

(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