Based on the following reaction, determine how much heat is released if 500. grams of methane is combusted in excess oxygen: CH4 + 2 O2  CO2 + 2 H2O H= – 890 kJ

- 13,150.6kJ

Explanation:

CH4 + 2 O2 > CO2 + 2 H2O ΔH= – 890 kJ

The ΔH is enthalpy change of combustion , which is the heat is either absorbed or released by the combustion of one mole of a substance.

ΔH=−890 kJ/mol (released in the combustion of one mole of methane)

using the molar mass  (in grams )of methane to get moles of sample

(237g × 1 mole of CH4)/16.04g=14.776 moles  of CH4

Since 1 mole produces 890 kJ of heat upon combustion, then 14.776 moles will produce

ΔH = 14.776moles of CH4 × 890kJ/1mole of CH4

=13,150.6kJ

Therefore ΔH = - 13,150.6kJ

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

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2