a. How many kilojoules are released when 75.0 g of CH4 O reacts? 2 CH4 O(l)+3 O2(g) ⟶2 CO2(g)+4

step 1 This question has two parts. The first part, Part A, is asking us to determine how many kilojoul

A. How many kilojoules are released when 75.0 g of CH4 O...

a. How many kilojoules are released when $75.0 \mathrm{~g}$ of $\mathrm{CH}_{4} \mathrm{O}$ reacts? $2 \mathrm{CH}_{4} \mathrm{O}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$ $\Delta H=-726 \mathrm{~kJ}$ b. How many kilojoules are absorbed when $315 \mathrm{~g}$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ reacts? $$ \mathrm{Ca}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=+65.3 \mathrm{~kJ} $$

a. How many kilojoules are released when $75.0 \mathrm{~g}$ of $\mathrm{CH}_{4} \mathrm{O}$ reacts?
$2 \mathrm{CH}_{4} \mathrm{O}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)$
$\Delta H=-726 \mathrm{~kJ}$
b. How many kilojoules are absorbed when $315 \mathrm{~g}$ of $\mathrm{Ca}(\mathrm{OH})_{2}$ reacts?
$$
\mathrm{Ca}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=+65.3 \mathrm{~kJ}
$$

Key Concepts

Stoichiometry

Stoichiometry involves using the balanced chemical equation to relate the amounts of reactants and products in terms of moles. This key concept allows one to convert between the mass of a substance and the number of moles using its molar mass, and then apply the mole ratios provided by the balanced equation to determine the corresponding amounts of energy involved in a reaction.

Enthalpy Change

Enthalpy change, symbolized as ?H, is the measure of heat absorbed or released during a chemical reaction at constant pressure. It indicates whether a reaction is exothermic (negative ?H, releasing heat) or endothermic (positive ?H, absorbing heat) and is essential for understanding the energy changes involved in chemical processes.

Molar Mass

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole, and serves as the conversion factor between the mass of a substance and the number of moles. This concept is crucial for connecting the quantitative aspects of a chemical reaction with the stoichiometric coefficients in the balanced chemical equation.

Transcript

00:01 This question has two parts.

00:02 The first part, part a, is asking us to determine how many kilojoules of heat are released when we have 75 .0 grams of methanol reacting.

00:13 So, what do we do? well, we have to convert the 75 .0 grams of methanol to moles.

00:22 So how many moles is 75 grams of methanol? well, to do this, we have to set up our conversion ratio here.

00:32 So, what's our conversion factor? well, we have to put down moles on the top, and then figure out how many grams are in a mole, specifically for methanol.

00:42 So methanol has the formula, c -h3 -o -h.

00:46 Okay? we know that c is 12.

00:50 H is going to be 1, and we have 4 of them.

00:56 So we're going to add in 4, and then we're going to add in 16 for the 0.

01:00 And that's 32.

01:04 So, it's going to be 75 grams divided by 32 grams in the mole.

01:12 So what do we get? 75 by 32 gives us 2 .34 moles of methanol.

01:27 So now we know that 2 moles of methanol are required to produce negative 726 kilojoules.

01:33 So what if we have 2 .34? well, we would basically figure out what we're going to multiply by, so 2 .34.

01:40 We're going to divide it by the 2 moles that are needed.

01:43 This is going to tell us how many, what's our multiplication factor here? so divided by 2, we get 1 .17.

01:53 All right.

01:54 So we take our negative 726 kilojoules, and we multiply it by our factor, which is 1 .17.

02:03 Because rather than having 2 moles reacting, like it says in the reaction, we actually have this extra 2 .34.

02:10 So we multiply.

02:12 What do we get negative 726 times 1 .17 gives us negative 849 kilojoules.

02:29 That's your answer to part a...

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