Given that the molar mass of h2o is 18.02 g/mol, how many liters of propane are required at stp to produce 75 g of h2o from this reaction?

Given:

H2O, mass 75g of H2O at STP

molar mass of H2O is 18.02 g/mol

Required:

Liters of propane

Solution:

C3H8 + 5O2 -> 3CO2 + 4H2O

get the number of moles of H2O using the molar mass of H2O is 18.02 g/mol

n = 75g/18.02 g/mol = 4.162 moles H2O

change moles H2O to moles propane

4.162 moles H2O(1 mole C3H8/4 moles H2O) = 1.041 moles C3H8

From the ideal gas equation PV = nRT

V = nRT/P

V = (1.041 moles C3H8)(0.08206 L-atm/mol-K) (273K)/(1 atm)

V = 23.3 L of C3H8

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

they are both synthetic elements

explanation:

d