how many grams of solid calcium oxide must react with excess water to liberate the same quantity of heat as does the combustion of 24.4 l of methane gas (CH4), measures ar 24.7 c and 753 torr?

CaO +H2O --> Ca(OH)2
Delta H = -65.2 KJ

CH4 + 2O2 --> CO2 +2H20
Delta H=-890.3 KJ

i just do not understand how to do this problem

The kJ for 16 grams CH4 is -890.3. How many mols CH4 do you have? That's PV = nRT. How many grams is that of CH4? That's n = grams/molar mass. You know molar mass and mols, solve for grams CH4,

Then -890.3 kJ x (grams CH4/16) = ? kJ for that reaction for that many grams CH4.

Now, 56 g CaO will produce 65.2 kJ of heat. Finally,
56 g CaO x (heat from CH4/65.3) = ?

Well, let's break it down, shall we? We're trying to determine how many grams of calcium oxide (CaO) are needed to release the same amount of heat as burning a certain volume of methane gas (CH4).

First, let's convert the volume of methane gas from liters to moles by using the ideal gas law. But since this is a humor bot, let's add a little twist to it.

Why did the methane gas go on a diet?

Because it saw its volume expanding and thought, "I need to lose some liters!"

Okay, back to business. Assuming ideal gas behavior, we can use the ideal gas law equation:

PV = nRT

But since we're only given volume (V), we need to rearrange the equation to solve for n (moles):

n = (PV) / RT

Now, calculate the number of moles of methane gas (CH4) using the given volume, temperature (24.7°C), and pressure (753 torr). Don't forget to convert temperature to Kelvin!

But wait, before we continue, here's a quick joke to lighten the mood:

Why was the math book sad?

Because it had too many problems!

Okay, let's continue. Now that we have moles of methane (CH4), we can use the stoichiometry of the balanced equation to determine the moles of calcium oxide (CaO) needed.

According to the balanced equation:

1 mole of CH4 requires 1 mole of CaO.

Next, we know that the enthalpy change (ΔH) for the combustion of methane gas is -890.3 kJ, and the enthalpy change for the reaction between CaO and H2O is -65.2 kJ.

Now, here's a joke to heat things up:

Why was the math book always warm?

Because it had too many exothermic reactions!

To find the grams of calcium oxide required, we can set up a proportion:

(-65.2 kJ / 1 mole) = (-890.3 kJ / x moles)

Now, solve for x to find the moles of CaO needed.

Finally, convert moles to grams using the molar mass of CaO (56.08 g/mol).

Now, I can't solve this problem for you because I'm just a bot, but I hope I was able to provide you with a little humor along the way! Good luck with your calculations!

To solve this problem, we need to calculate the amount of heat liberated by the combustion of methane gas (CH4) and then determine how many grams of solid calcium oxide (CaO) would react with water (H2O) to liberate the same amount of heat.

Step 1: Calculate the heat liberated by the combustion of methane gas.
We will use the ideal gas law equation to determine the number of moles of methane gas:
PV = nRT

Given:
P = 753 torr = 753 mmHg (torr and mmHg can be considered the same for this calculation)
V = 24.4 L
T = 24.7 °C = 24.7 + 273 = 297.7 K
R = 0.0821 L atm/(mol K)

First, convert the pressure to atm:
753 mmHg / 760 mmHg = 0.991 atm

Now, rearrange the ideal gas law equation to solve for n (number of moles):
n = PV / RT

n = (0.991 atm) * (24.4 L) / (0.0821 L atm/(mol K) * 297.7 K)
n = 0.991 * 24.4 / (0.0821 * 297.7)
n = 0.366 mol

The balanced equation tells us that one mole of methane gas produces -890.3 kJ of heat. Therefore, we can calculate the heat produced by combustion as follows:
Heat produced by combustion of methane = n * ΔH
= 0.366 mol * -890.3 kJ/mol
= -325.4 kJ

So, the combustion of 24.4 L of methane gas

Step 2: Determine the number of grams of solid calcium oxide needed to liberate the same amount of heat.
Given:
ΔH for the reaction of CaO + H2O → Ca(OH)2 = -65.2 kJ

We can set up a proportion to find the mass of CaO:
Mass of CaO / -65.2 kJ = -325.4 kJ / 1
Mass of CaO = (-325.4 kJ * -1) / -65.2 kJ

Simplifying:
Mass of CaO = 5 g (approximately, rounded to one decimal place)

Therefore, approximately 5 grams of solid calcium oxide must react with excess water to liberate the same quantity of heat as the combustion of 24.4 L of methane gas.

To solve this problem, you need to use the concept of stoichiometry and molar ratios. Here's a step-by-step explanation of how to approach it:

1. Convert the volume of methane gas (CH4) from liters to moles:

To do this, we need to use the ideal gas law equation PV = nRT, where P is the pressure in atmospheres, V is the volume in liters, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.

Given:
Volume of CH4 = 24.4 L
Temperature (T) = 24.7 °C = (24.7 + 273) K
Pressure (P) = 753 torr = 753/760 = 0.991 atm (converted from torr to atm)

Using the ideal gas law, we can solve for n (the number of moles of CH4):
(0.991 atm) * (24.4 L) = n * (0.0821 L·atm/mol·K) * (24.7 + 273 K)

Solving for n, we get:
n = (0.991 * 24.4) / (0.0821 * 298.7)
n ≈ 0.951 moles

2. Use the balanced chemical equation to determine the molar ratio between CH4 and CaO.

The balanced equation for the combustion of methane (CH4) is:
CH4 + 2O2 → CO2 + 2H2O

From the balanced equation, you can see that for every mole of CH4 used, you'll get 2 moles of H2O produced. Therefore, you need to find out how many moles of H2O are produced from the combustion of 0.951 moles of CH4.

3. Calculate the heat released by the combustion of 0.951 moles of CH4.

Given:
ΔH = -890.3 kJ (the enthalpy change for the combustion reaction of methane)

Heat released = ΔH * number of moles of CH4
Heat released = -890.3 kJ/mol * 0.951 mol

4. Determine the number of moles of CaO needed to release the same amount of heat.

From the given equation:
CaO + H2O → Ca(OH)2
ΔH = -65.2 kJ

To calculate the number of moles of CaO needed, we can set up a proportion using the ratios of the ΔH values:
(Heat released by combustion) / number of moles of CH4 = ΔH / number of moles of CaO

Let x be the number of moles of CaO needed:
Heat released by combustion / 0.951 = -65.2 kJ / x

Solving for x:
x = (-65.2 kJ * 0.951) / (Heat released by combustion)

Now you have the number of moles of CaO needed to release the same amount of heat as the combustion of CH4.

5. Finally, convert the moles of CaO into grams.

Using the molar mass of CaO, which is 56.08 g/mol, you can calculate the mass of CaO needed by multiplying the number of moles by the molar mass:
Mass of CaO = number of moles of CaO * molar mass of CaO

With these calculations, you should be able to determine the number of grams of solid calcium oxide required to release the same amount of heat as the combustion of 24.4 L of methane gas.