The steps for balancing equations are as follows and should be done in order:

1.) Check for diatomic molecules: In this case there are none so move on to step 2

2.) Balance the metals (hydrogen is not considered a metal in this application): We have no metals in our equation, so move on to step 3 (the sodium is part of the catalyst so we don't consider it part of the balancing). 

3.) Balance the nonmetals (not including oxygen). If you notice the left hand side of the equation has 3R groups and the right hand side only has one R-group. To fix this we can put a coefficient of 3 in front of  on the right hand side giving us the following:

On the left hand side of the equation we have 7 carbon atoms but on the right hand side we have 9 carbon atoms. This can be rectified by putting a coefficient of 3 in front of  on the left hand side of the equation giving us the following:

Now all the nonmetals (excluding oxygen) have been balanced and we can move on to step 4

4.) Balance oxygen. There are 9 oxygen atoms on both sides of the equation. Move on to step 5

5.) Balance Hydrogen. There are 17 hydrogens on both sides of the equation.

6.) Recount all atoms to make sure you have balanced correctly. 9 carbons on both sides, 3 R-groups on both sides, 9 oxygen on both sides, and 17 hydrogens on both sides.

If this problem had involved ionic species you would also want to make certain that the charges are balanced on both sides of the equation.

The charges on both sides add up to zero so your equation is balanced:

To balance chemical reactions, it's usually more effective to save the more common elements, such as oxygen, for last. For this reaction, let's go ahead and start by balancing the element  on both sides.

Because there are  moles of  on the right side, we'll need to have  moles of . Next, let's go ahead and balance nitrogen on both sides.

Since we added a coefficient of  to the  in the first step, there are now  moles of  on the left. Thus, we'll need to add a coefficient of  to the . Next, if we check the amount of , , and  on both sides, we see that they are equal. Thus, we can recognize that  and  have coefficients of , and do not need to be altered.

Our final balanced equation looks like this.

When considering Limiting Reactant problems the most important aspect to consider is the molar ratio of the reactants.  Here the balanced formula tells us that for every 2 moles of Ca there must be 1 mole of O₂ to create the product.  The amounts given by the problem are the actual amounts we are given and can be compared to the molar ratio to determine the limiting reactant.  Since there is 1 mole of O_2, in any situation in which there are less than 2 moles of Ca, Ca is the limting reactant.  Conversely if there were 2 moles of Ca, any situation in which there less than 1 mole of O₂ would be a situation in which O₂ is the limiting reactant.  

H₂SO₄ molecular weight is 98. This gives 1.02moles. Since only 1 mole is needed per reaction it allows the reaction to go through 1.02 times.

LiOH molecular weight is 24. This gives 2.71 moles. Every reaction requires 2 moles so the reaction can go through 1.35 times. 

Based on this, H₂SO₄ is the limiting reactant.

In irreversible reactions, the reaction proceeds in one direction only and these reactions usually go to completion. Since the forward reaction is the only one that occurs, the amount of product formed is only based on the reactants present, namely, the amount of limiting reactant. 

First, calculate the theoretical yield as if we had 293 g and excess :

Then calculate the theoretical yield as if we had 17.2 g :

The limiting reagent is as the theoretical yield calculation is lower. Now all there is to do is convert moles into grams:

First, let us consider the 36 grams of solid sodium

We now know that sodium is the limiting reagent as it uses only 28.2 mL of water and we have 53 mL of water. We will have no remaining sodium.

We will now calculate the remaining volume of water

The reactant that will be depleted first is called the limiting reagent. We can determine the limiting reagent by calculating how much oxygen gas is necessary to use up all 60 grams of magnesium.

We start by converting the magnesium to moles.

We then compare the molar ratio of magnesium to oxygen gas. Since we need 1 mol of oxygen gas for every 2 moles of magnesium, we only need 1.24 moles of oxygen gas to fully react the magnesium.

Multiplying by the molar mass, we determine that we need 39.51 grams of oxygen gas. Remeber that the oxygen is diatomic in its gaseous state.

Since we only have 30 grams in the givevn question, we conclude that oxygen gas will be depleted first if the reaction runs to completion.

When reading the balanced reaction, you should notice that it requires 1 mol of chlorine gas and 2 moles of potassium iodide in order to react and create the products. This ratio is seen in the coefficients of the reagents.

Using this ratio, we can calculate the mole of chlorine needed to react with 10 moles of potassium iodide.

5 moles of chlorine gas will be used to react all of the potassium iodide. Since we started with 8 moles of chlorine gas, this means that there are 3 moles of chlorine gas left over after the reaction has run to completion. 

Since we are given masses in this problem, it is a little trickier to compare the amounts of both reagents to one another. We can use stoichiometry to determine how much of one reagent is needed in order to use up all of the other reagent. We can determine how much potassium iodide is necessary to use up all of the chlorine gas by using the following series of calculations.

Convert both compounds to moles.

Then, use the ratio in the chemical reaction to determine the moles of potassium iodide needed to react all of the chlorine.

Only 2.82 moles of poatssium iodide are needed. Since we started with 3.01 moles of potassium iodide, 0.19 moles will be left over. Now, we need to convert this remainsing amount to grams.