What mass of propane could burn in 48.0 g of oxygen?

The balanced combustion reaction of propane, C₃H₈, is

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

Molar mass of propane: 44 g/mol

Moles of propane = 42 g * (1 mol/44g) = 0.9545 mol propane

Molar mass of oxygen: 32 g/mol

Moles of oxygen = 115 g * (1 mol/32 g) = 3.594 mol oxygen

Moles of oxygen needed to completely react propane:

0.9545 mol propane * (5 mol O₂/1 mol propane) = 4.7725 mol oxygen

Since the available oxygen is only 3.594 moles and propane needs 4.7725 moles, that means oxygen is our limiting reactant. We base the amount of water produced here.

Molar mass of water: 18 g/mol

Mass of water produced = 3.594 mol O₂ * (4 mol H₂O/5 mol O₂) * (18 g/mol)

Mass of water produced = 258.768 grams

Explanation

am i too late?

explanation:

(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