Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
white phosphorus is used in flares and explosives, so may be important in warfare, etc. Red phosphorus is used in matches (side of matches) and in fertilizers which are essential to growing plants. I don't know if that is economically important, but there are many ways that phosphorus benifit the economy. One downside on the economy is that phosphorus is one of the main ingredients in meth, which of course causes trouble for the economy ( law enforcement, medical, etc. )
a. volume of NO : 41.785 L
b. mass of H2O : 18 g
c. volume of O2 : 9.52 L
<h3>Further explanation</h3>
Given
Reaction
4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)
Required
a. volume of NO
b. mass of H2O
c. volume of O2
Solution
Assume reactants at STP(0 C, 1 atm)
Products at 1000 C (1273 K)and 1 atm
a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :
volume NO at 1273 K and 1 atm
b. 15 L NH3 at STP ( 1mol = 22.4 L)
mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :
mass H2O(MW = 18 g/mol) :
c. mol NO at 1273 K and 1 atm :
mol ratio of NO : O2 = 4 : 5, so mol O2 :
Volume O2 at STP :
