We need to balance the chemical equation first. Count the numbers of atoms that come in with the reactants, and compare that to those consumed by the products. I like to do this with pencil and paper so that you can make changes easily until you get something that works.
I'll assume you know the basics of balancing. If not, I have a more detailed explanation here: https://drive.google.com/file/d/1CkKeVpdNTTwd95TqKT_taFSCMSX8nE6z/view?usp=sharing
The example is a different equation, but the same principles apply. The important part is to stay organized and start with the most complex molecule first. In this case it is the pentane. You need at least one molecule of each compound in the equation, so pencil in a "1" for the pentane. From that, you see that you need 5 carbons. The only place carbons can go is the CO2, so jot down a "5" in front of it.
The pentane brings in 12 hydrogens. They can only go to the H2O, so put a "6" in front of H2O. Both of the products have oxygen, which can only come from the O2. We already have guessed the numbers of CO2 and H2O, so add the oxygens there, which comes to 16 total. To get 16 oxygens, we need 8 O2 molecules. And after double-checking, we can see the equation is now balanced.
We want to know how many grams of H2O will be produced with 40 grams of pentane. Our equation says that we'll get 6 moles of H2O for every 1 mole of C5H12. So let's calculate the number of moles of pentane.
Moles = mass of pentane/molar mass of pentane. Add the atomic weights of each atom in the molecule. For pentane that is 72.1 AMU. Then simply rewrite the same number as 72.1 grams pentane/mole pentane. Part of the power of the periodic table is the ease for calculating a number that can be used measure out numbers of molecules by their mass, instead of counting them. One mole of anything is equal to Avogadro's number (6.02 x 1023 ). So if we measure out 72.1 grams of pentane, we'll have that number of pentane molecules. This equation tells us that for that one mole coming in, we'll get 6 moles of H2O, or 3.61 x 1024 molecules of H2O coming out
40 grams of pentane = 40 grams pentane/(72.1 grams pentane/mole pentane). This calculates to 0.555 moles of pentane (3 sig figs). Assuming it all reacts with the excess oxygen, we should get 6 times that number of moles of H2O. Therefore, ((6moles H2O)/(1 mole pentane))*(0.555 moles pentane) = moles of H2o produced. I calculate that to be 3.33 moles H2O.
They want the answer in grams H2O, so convert moles to grams using the molar mass of water (18 g/mole).
3.33 moles H2O * 18 g H2O/mole H2O = 19.9 grams H2O
I hope that helps. Use the same procedure to calculate the second problem. Start with the most complex molecukle when balancing, which I judge to be the Li2CO3. You can see that you need at least 2 Li on the reactant side, if you start with 1 molecule of Li2CO3 on the product side.
Once it is balanced, convert the lithium hydroxide into moles and then use the balanced equation to calculate how many moles of lithium carbonate will be produced, assuming excess CO2.