Butane, c4h10, is a component of natural gas that is used as fuel for cigarette lighters. the balanced equation of the complete combustion of butane is 2c4h10(g)+13o2(g)→8co2(g)+10h2o(l) at 1.00 atm and 23 ∘c, what is the volume of carbon dioxide formed by the combustion of 1.60 g of butane?

Reaction of combustion of butane is a as follows:

2 C₄H₁₀ + 13 O₂ --> 8 CO₂ + 10 H₂O:  

As it can be seen from the equation that 2 mole of butane produce= 8 mole of butane produce  

1 mole of butane produce= 4 mole of butane produce

1.60 g of butane=

= 0.027 mole of butane

0.027 mole of butane =

                                     = 0.108 mole of CO₂.

Using ideal gas equation,  

Here,  

P denotes pressure  

V denotes volume  

n denotes number of moles of gas  

R denotes gas constant  

T denotes temperature  

The values at STP will be:  

P=1 atm  

T=23+273 K = 296 K

R=0.0821 atm L mol ⁻¹

Number of moles of gas, n= 0.018

Putting all the values in the above equation,

V= 2.62 L

So the volume will be 2.62 L

the electrical  conductivity  of asolution  of an electrolyte is measured by determining the resistance of thesolution  between two flat or cylindrical electrodes separated by a fixed distance.

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