Practice: Determine the products, balance the equation, and identify as organic or inorganic combustion reaction (circle).
1)
Ca + O2
Inorganic or Organic combustion?
2)
C3H2O +
02
Inorganic or Organic combustion?

1. Inorganic combustion

2. Organic combustion

Given

The chemical equations

Required

Balanced equation

Solution

Organic compounds have the characteristic that there is a chain of carbon atoms, while inorganic does not have a chain of carbon atoms

The reaction of a substance with oxygen is called a combustion reaction

Complete combustion of carbon compounds will get CO₂ gas, whereas if it is not complete it will produce CO gas

So Organic : contain a carbon atom (and often a hydrogen atom)

1. Ca + O₂ ⇒ CaO₂

Inorganic combustion⇒does not contain carbon

2. C₃H₂O + 3O₂ → 3CO₂ + H₂O

Organic combustion

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol