Question

Consider the following thermochemical equation:

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g) ∆Horxn = –1277 kJ
What mass of water can be heated from 23.5°C to 88.2°C with the heat that is released from the combustion of 12.0 g of CH3OH (molar mass = 32.0 g/mol), assuming no heat is lost to the surroundings? Specific heat capacity of H2O = 4.184 J/g°C.

Answer 1

`m_w=5.3248* 10^(23)\ kg`

Step-by-step explanation:

According to the given reaction,

heat released by thecombustion of 2 molecules of Methanol,
`\Delta H=1277000\ J`

we know that molecular mass of Methanol,
`m=32\ g.mol^(-1)`

∴12 gram of methanol =
`(3)/(8)\ moles`

we know 1 mole =
`6.023* 10^(23)\ molecules`

so,

`\rm (3)/(8) \ moles=2.258* 10^(23)\ molecules`

Heat from the combustion of
`2.258* 10^(23)\ molecules`:

`Q=1277000* (2.258* 10^(23))/(2)`

`Q=1442132.0625* 10^(23)\ J`

Now the mass of water that can be heated from 23.5°C to 88.2°C :

`Q=m_w.c_w.\Delta T`

`1442132.0625* 10^(23)=m_w.4186* (88.2-23.5)`

`m_w=5.3248* 10^(23)\ kg`