What volume of carbon dioxide gas at STP is required to produce 456.9g of lithium carbonate? PLEASE EXPLAIN HOW YOU GOT THE ANSWER

V =  138.68 L

Explanation:

Given data:

Volume of CO₂ at STP required = ?

Mass of lithium carbonate produced = 456.9 g

Solution;

Chemical equation:

2LiOH + CO₂     →     Li₂CO₃ + H₂O

Number of moles of Li₂CO₃ produced:

Number of moles = mass/molar mass

Number of moles = 456.9 g/ 73.89 g/mol

Number of moles = 6.184 mol

now we will compare the moles of Li₂CO₃ with CO₂ from balanced chemical equation to get number of moles of CO₂  reacted.

                      Li₂CO₃      :        CO₂

                           1           :          1

                         6.184     :        6.184

Thus, 6.184 moles of  CO₂ reacted.

Volume of CO₂:

PV = nRT

R = general gas constant = 0.0821 atm.L /mol.K

1 atm × V = 6.184 mol ×0.0821 atm.L /mol.K × 273.15 K

V = 138.68 atm.L / 1 atm

V =  138.68 L

(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