First, we must determine how many moles of each reactant begin the reaction by multiplying the molarity by the volume. Don't forget to convert volume to liters!

Next, use the reaction coefficients (i.e. the stoichiometry) to determine how many moles of calcium carbonate could be formed from each of the reactants. In this case, there is a 1:1 molar ratio between both reactants and calcium carbonate.

Thus, 0.0125 moles of potassium carbonate could form 0.0125 moles of calcium carbonate, while 0.0175 moles of calcium nitrate could form 0.0175 moles of calcium carbonate. The maximum amount of product is going to be determined by the limiting reactant, i.e. the reactant that provides the least amount of product. In this case, the limiting reactant is potassium carbonate, and the maximum yield of calcium carbonate is 0.0125mol.

For the final step convert this value to grams:

When two or more reactants of a chemical equation's quantities are given, the first step is to determine the limiting reagent of the reaction. The reaction will only proceed and produce as much product as the limiting reagent allows. For example if there are 5 moles of oxygen gas and excess hydrogen gas, only 10 moles of water can be produced regardless of how much hydrogen gas is present. 

First of all, a product cannot be the limiting reactant. This problem can be solved in many different ways. Keep in mind that to solve a stoichiometry problem a good practice is to convert mass to mol. Let's find if the amount of  we have is enough to react with  of :

We have only  of   and  are needed. Then  is the limiting reactant.

Soot is a mixture incompletely combusted hydrocarbons. Therefore the production of soot occurs most when the hydrocarbon is in excess and the oxidizer (in this case oxygen) is the limiting reactant.

Incorrect answers and explanations:

The limiting reactant is propane, therefore the best way to reduce the soot is to increase the input of propane.  

This is incorrect because propane is in excess when soot production is maximized. Therefore adding more propane will only result in more soot.

The limiting reactant is oxygen; therefore the best way to reduce the soot is to add an oxygen source such as , which releases oxygen by thermal decomposition.

The limiting reactant is indeed oxygen and  does undergo thermal decomposition to produce oxygen. However it is also one of the primary ingredients for gunpowder (the other two ingredients have a high probability of being present in a device that processes tissue from living organisms), therefore the safety factor of this answer leads it to be incorrect. Not to mention that this method would not be the simplest solution.

Oxygen is in excess; therefore the best way to reduce the soot is to close the grill hood and close any vent holes preventing atmospheric oxygen from entering the system.

Oxygen is the limiting reactant; therefore cutting off the supply of oxygen would only create more soot or depending upon the degree to which the oxygen supply is cut off, this could stop the reaction altogether.

The soot production is due to a dirty grill and therefore your grill must be thoughly cleaned.  

A dirty grill may have soot in it, but unless the dirtiness is blocking off the supply of oxygen, the soot currently being produced is not caused by the dirt. In any case the root cause of any soot production is that oxygen is limiting and this answer does not address that fact.

Correct Answer and explanation:

The limiting reactant is oxygen; therefore the best way to reduce the soot is to decrease the input of propane, which is in excess.

The limiting reactant when soot is produced is oxygen. In most propane grills, the oxygen supply is maxed out at the level of atmospheric oxygen. Therefore the only reactant under your direct control is the propane. By reducing the quantity of propane, you can make it no longer in excess.

First we must determine the volume of the container:

The density is .

Now we will convert this into grams per milliliter.

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

The net ionic equation that occurs is:

 and  are spectator ions. Hence,   concentration is not affected. The molecular mass of  is , then  of NaCl yields  of . Since we have  of a  solution of , we have  of . There is no limiting reactant. Hence  of  are formed. Being the molecular mass of ,  are formed. Result should be given with tree significant figures.

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.

Remember, there really isn't a way to go straight from grams to atoms, but, it's possible to change grams into moles (using the molar mass), and it's possible to change moles into atoms (using Avogadro's number).

The first step is to determine the molecular weight of the molecule:

The molecular weight serves as a conversion factor to convert grams to moles as implied its units . So, given that we are dealing with 20.0 grams, the moles of 20.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

Based on the molecular formula, for every mole of the compound , there is 1 mole of bromine. Based on our calculations 20.0 grams of  is 0.1835 moles of the compound. Because there is a 1:1 molar ratio of bromine to every molecule, we have 0.1835 moles of bromine in 20 grams of our compound, .