1.) Consider the reaction: B2H6(g) + 3O2(g) àB2O3(s) + 3H2O(g) H = -2035 kJ Calc
ID: 926776 • Letter: 1
Question
1.) Consider the reaction:
B2H6(g) + 3O2(g) àB2O3(s) + 3H2O(g) H = -2035 kJ
Calculate the amount of heat released when each of the following amounts of diborane (B2H6) is burned.
1.0 g
2. Consider the reaction:
B2H6(g) + 3O2(g) àB2O3(s) + 3H2O(g) H = -2035 kJ
Calculate the amount of heat released when each of the following amounts of diborane (B2H6) is burned.
1.0 mol
3. Consider the reaction:
B2H6(g) + 3O2(g) àB2O3(s) + 3H2O(g) H = -2035 kJ
Calculate the amount of heat released when each of the following amounts of diborane (B2H6) is burned.
1.0 x 10 2 mol
4. Consider the reaction:
B2H6(g) + 3O2(g) àB2O3(s) + 3H2O(g) H = -2035 kJ
Calculate the amount of heat released when each of the following amounts of diborane (B2H6) is burned.
A mixture of 10.0 g B2H6 and 10.0 g O2
5. A 15.0 g sample of nickel metal is heated to 100.°C and dropped into 55.0 g of water,
initially at 23.0°C. Assuming that all of the heat lost by the nickel is absorbed by the
water, calculate the final temperature of the nickel and the water. The specific heat of
nickel is 0.444J/g°C)
6. Given the following data:
C2H2(g) + 5/2 O2(g) à2CO2(g) + H2O(l) H= -1300.kJ
C(s) + O2(g) àCO2(g) H= -394kJ
H2(g) + O2(g) à H2O(l) H= -286kJ
Use Hess’s Law to calculate H for the reaction
2C(s) + H2(g) àC2H2(g)
7. Calculate H° for each of the following reactions using the data for the enthalpy of
formation in the appendix.
2Na(s) + 2H2O(l)à2NaOH(aq) + H2(g)
Explanation / Answer
1. B2H6(g) + 3O2(g) ---> B2O3(s) + 3H2O(g) H = -2035 kJ
This means 2035 kJ of energy is released when 1 mole of B2H6 gas is reacted with oxygen.
Lets convert 1g of B2H6 into moles
Number of moles = weight/molecular weight = 1/27.67 =0.036 moles
Energy released = 0.036 * 2035 = 73.55 kJ
2) As mentioned above, according to given equation 2035 kJ of energy is released when 1 mole of B2H6 gas is burned.
3) Energy released = number moles* 2035 kJ =100 * 2035 kJ= 203500 kJ=2035 X 102 kJ
4) A mixture of 10.0 g B2H6 and 10.0 g O2
10 g of B2H6= 10/27.67 =0.362 moles
10.0 g O2 = 10/32 =0.3125 moles
According given equation the molar ratio between B2H6 and O2 is 1:3.
So, here O2 is a limiting agent and 0.362/3 = 0.120 moles of B2H6 will undergo combustion.
Energy released = number moles* 2035 kJ = 0.120 * 2035 kJ = 245.56 kJ
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.