Question

Can you help me this two parts of the question ?

Answer 1

`379.8\text{ g}`

Step-by-step explanation:

Here, we want to get the mass of methylamine that must be combusted

From the question, we have the following:

`\begin{gathered} \Delta T=\text{ 51-19 = 32\degree C} \\ mass\text{ of water = 395g} \\ change\text{ in enthalpy of formation:} \\ water\text{ = -285.8} \\ Methylamine\text{ = -23.0} \\ Carbon\text{ }(iv)\text{ oxide = -393.5} \end{gathered}`

Now, we write the equation of reaction as follows:

`4CH_3NH_2\text{ + 9O}_2\text{ }\rightarrow\text{ 4CO}_2\text{ + 10H}_2O\text{ +}2N_2`

We have the overall change in enthalpy as follows:

`\begin{gathered} \Delta H\text{ = }\Delta H_(products)\text{ - }\Delta H_(reactants) \\ \Delta H\text{ = 4}(-393.5)\text{ + 10}(-285.8)\text{ -}(4*(-23.0)) \\ \Delta H\text{ = -4340 KJ/mol} \end{gathered}`

What this simply means is that 4340 KJ/mol of heat is released when 4 moles of methylamine is burned

Mathematically:

`Q\text{ = mc}\Delta\theta`

m is the mass of the methylamine we want to calculate

c is the specific heat capacity (For water, it is 4.2)

change in temperature is delta theta

What we want to calculate here is the amount of heat gained by the water

We have that as:

`Q\text{ = 395 }*4.2*\text{ 32 = 53,088 KJ}`

Now, we proceed to get the number of moles of methylamine burned actually

That would be:

`(Q)/(\Delta H)\text{ = }(53,088)/(4340)\text{ = 12.23 mol}`

Finally, to get the mass, we multiply the number of moles by the molar mass

The molar mass of methylamine is 31.05 g/mol

Thus, we have the mass as:

`31.05*12.23\text{ = 379.8g}`