Calculate the amount of heat needed to melt 89.9 g of solid methanol (CH3OH) and bring it to a temperature of 49.1C.

20075.35 J

Explanation:

We'll begin by calculating the number of mole in 89.9 g of CH₃OH. This can be obtained as follow:

Mass of CH₃OH = 89.9 g

Molar mass of CH₃OH = 12 + (3×1) + 16 + 1

= 12 + 3 + 16 + 1

= 32 g/mol

Mole of CH₃OH =?

Mole = mass / molar mass

Mole of CH₃OH = 89.9 / 32

Mole of CH₃OH = 2.81 moles

Next, we shall determine the heat required to melt the solid methanol (CH₃OH). This can be obtained as follow:

Mole of CH₃OH (n) = 2.81 moles

Heat of fusion (Hբ) = 3.17 KJ/mol

Heat required to melt CH₃OH (Q₁) = ?

Q₁ = n × Hբ

Q₁ = 2.81 × 3.17

Q₁ = 8.9077 KJ

Converting to J

Q₁ = 8.9077 × 1000

Q₁ = 8907.7 J

Next, we shall determine the heat required to change the temperature of methanol to 49.1 °C. This can be obtained as follow:

Mass of CH₃OH (M) = 89.9 g

Initial temperature (T₁) = 0 °C

Final temperature (T₂) = 49.1 °C.

Specific heat capacity of CH₃OH (C) = 2.53 J/gºC

Heat required to change the temperature (Q₂) =?

Q₂ = MC(T₂ – T₁)

Q₂ = 89.9 × 2.53 × (49.1 – 0)

Q₂ = 89.9 × 2.53 × 49.1

Q₂ = 11167.65 J

Finally, we shall determine the total heat. This can be obtained as follow:

Heat required to melt CH₃OH (Q₁) = 8907.7 J

Heat required to change the temperature (Q₂) = 11167.65 J

Total heat required (Q) =?

Q = Q₁ + Q₂

Q = 8907.7 + 11167.65

Q = 20075.35 J

Therefore, the total heat required to melt the methanol and bring it to a temperature of 49.1 °C is 20075.35 J

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.