A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature rose by 5.72°C. If the heat capacity of the bomb plus water was 9.47 kJ / °C, calculate the molar heat of combustion of methanol.

872.28 kJ/mol

Explanation:

The heat released is:

ΔH = C*ΔT

where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:

ΔH =  9.47*5.72 = 54.1684kJ

A quantity of 1.922 g of methanol in moles are:

moles = mass / molar mass

moles = 1.992/32.04 = 0.0621 mol

Then the molar heat of combustion of methanol is:

ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g