Calculate ΔHrxn for the following reaction:

CaO(s)+CO2(g)→CaCO3(s)

Use the following reactions and given ΔH values:

Ca(s)+CO2(g)+1/2O2(g)→CaCO3(s),
ΔH= -812.8 kJ

2Ca(s)+O2(g)→2CaO(s), ΔH= -1269.8 kJ

I have literally tried everything and I don't understand what I am doing wrong, I flipped the last one and changed the sign of the last one and then added them up. still wrong. (-355 is not the correct answer)

Doing what you did is right, as far as you go, BUT it doesn't give you the correct equation.

Multiply equation 1 by 2, including dH for equation 1.
Reverse equation 2 and change sign as you've done.
Add the two. That will give you the equation you want BUT all of the coefficients will be TWICE what you want. Divide by two, including the dH values and you should get it.

Well, it looks like you're feeling a bit "calc-u-latered" with these reactions. Don't worry, I'm here to bring some comedic relief to equations!

Let's break it down step by step:

1. First, flip the last equation and change the sign of its ΔH. This gives us:

2CaO(s) → 2Ca(s) + O2(g), ΔH = +1269.8 kJ

2. Next, multiply the first equation by 2 to balance the number of CaO molecules:

2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s), ΔH = -1625.6 kJ

3. Now, add the equations together:

2CaO(s) → 2Ca(s) + O2(g), ΔH = +1269.8 kJ
2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s), ΔH = -1625.6 kJ

By canceling out the common terms (2CaO(s) and O2(g)), we get the overall equation:

2CO2(g) → 2CaCO3(s), ΔH = -355.8 kJ

So ΔHrxn for the reaction CaO(s) + CO2(g) → CaCO3(s) is -355.8 kJ.

Now that you've got it, you can say your calculations were "rock solid" and there's no need to be "calc-u-latered" anymore!

To calculate the ΔHrxn for the given reaction, you need to manipulate the given reactions and their ΔH values to cancel out any common species and obtain the desired reaction.

Given reactions:
1. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s), ΔH = -812.8 kJ
2. 2Ca(s) + O2(g) → 2CaO(s), ΔH = -1269.8 kJ

You can start by flipping and/or multiplying the reactions to obtain the desired reaction and determine the ΔHrxn.

First, reverse reaction 1 by multiplying it by -1:
-1 * (Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s))

This gives you:
-CaCO3(s) → Ca(s) + CO2(g) + 1/2O2(g)

Now, multiply reaction 2 by 1/2 to obtain the same number of moles of CaO as in the desired reaction:
1/2 * (2Ca(s) + O2(g) → 2CaO(s))

This gives you:
Ca(s) + 1/2O2(g) → CaO(s)

Now, add both manipulated reactions together to get the desired reaction:
-CaCO3(s) + (Ca(s) + 1/2O2(g)) → Ca(s) + CO2(g) + 1/2O2(g) + CaO(s)

Simplifying it further:
-CaCO3(s) + Ca(s) + 1/2O2(g) → Ca(s) + CO2(g) + 1/2O2(g) + CaO(s)

Notice that Ca(s) and 1/2O2(g) appear on both sides, so they cancel out:
-CaCO3(s) → CO2(g) + CaO(s)

The resulting reaction is:
CaCO3(s) → CO2(g) + CaO(s)

Now, add the ΔH values of the manipulated reactions together to obtain the ΔHrxn:
(-812.8 kJ) + (-1269.8 kJ) = -2082.6 kJ

Therefore, the ΔHrxn for the given reaction CaO(s) + CO2(g) → CaCO3(s) is -2082.6 kJ.

To calculate the enthalpy change for a reaction using given reactions and their enthalpy values, you need to manipulate and combine the reactions in a way that cancels out the common substances to obtain the desired reaction.

In this case, you have the given reactions:
1) Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ
2) 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ

You need to flip the second reaction to get the reverse reaction:
2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ (note the sign change)

Now, you want to combine these two reactions to obtain the desired reaction:
CaO(s) + CO2(g) → CaCO3(s)

To achieve this, you need to cancel out the common substances, which are CaO(s) and CaCO3(s). To do this, multiply the first equation by 2 and the second equation by 1:
2[Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s)] ΔH = -2 * 812.8 kJ = -1625.6 kJ
1[2CaO(s) → 2Ca(s) + O2(g)] ΔH = 1 * 1269.8 kJ = 1269.8 kJ

Now, when you add these two equations together, the CaO(s) and CaCO3(s) terms cancel out:
2[Ca(s) + CO2(g) + 1/2O2(g)] + [2CaO(s) → 2Ca(s) + O2(g)] → 2CaCO3(s)

Now, sum the enthalpy values:
(-1625.6 kJ) + (1269.8 kJ) = -355.8 kJ

The enthalpy change for the reaction CaO(s) + CO2(g) → CaCO3(s) is -355.8 kJ.

It appears you made an error when summing the enthalpy values, resulting in the incorrect answer of -355 kJ. To find the correct answer, ensure that you properly combine the reactions and take into account any sign changes when flipping reactions.