# Stoich Project

## Assigned reaction

Calcium Carbonate and Nitric Acid

## Type of Reaction

This is a double replacement reaction because we are starting out with two aqueous compounds. In a double replacement reaction one must replace two things that are the same state of matter. For example Calcium and Hydrogen would switch places because they are both metals or solids.

## Balanced Equation

Ca(CO)3(aq) + 2HNO3(aq) → (CO2)(g) + Ca(NO3)2(aq) + H2O(l)

Since this is a double replacement reaction you replace two elements that are of the same matter for example in this problem you would replace Calcium and Hydrogen because they are both metals.

## Name for each reactant and product

The way you differentiate between a product and reactant is by simply by looking at what is left of the yielding sign and the right. Whatever is to the left of the is the reactant because that is what you start out with and to the right is the product because that is what you end up with.

Ca(CO)3(aq) + 2HNO3(aq) → (CO2)(g) + Ca(NO3)2(aq) + H2O(l)

In this example Calcium Carbonate and Nitric acid are the reactants ,and Nitric Acid is the product.

## Molar Mass

The way one finds the molar mass is by looking at the atomic mass of each element which is the bottom number under each element in the periodic table. But in some instances you have to multiply the atomic number by the sufficient of the compound or element.

An example of this is my first reactant Ca(CO3) in order to find the molar mass i would find the molar mass of calcium which is its atomic number. Second i would find the molar mass of Carbonate to do so i would find the atomic number of carbon which is 12.011 grams and add that by the molar of oxygen. However, one must multiply the molar mass of oxygen three times then add it to the molar mass of carbon and Calcium. That being said the molar mass of Calcium Carbonate would be 100.086 grams.

## Mole to Mole conversions

In order to do a mole to mole conversion first one must start with what they are given then in order to cancel out what you start with you must put the same element on the bottom of the train track but in front of it write the coefficient to the balanced equation after this above one must put the coefficient of the compound that you are looking for.

Example Problem:

11.30 Ca CO3 * 1 Ca CO3 mole/1 CO2 moles= 11.3 moles CO2

In this problem You start out with 11.30 moles of Calcium Carbonate and are trying to get to Carbon dioxide once you set it up like i have all one must do is multiply across, and divide across the bottom your answer will be answer will be 11.30 moles CO2.

## Mass to Mass conversions

In these types of conversions one is starting out with a number of grams in an element and using that to convert to another.

Step 1: After you right your given grams of an compound you put its molar mass on the bottom.

Step 2: Put one mole of that same compound over the molar mass of that compound.

Step 3: Then put the coefficient of the element that you started out with and put that under the coefficient that you are looking for.

Step 4: After step 3 is completed put 1 mole of whatever your looking for under the molar mass of whatever it is.

Example:

9.24 g Ca CO3 * 44.01 g CO2 /100.86 g Ca CO3=4.031 g CO2

## Limiting and Excess reactant

The way one must find the limiting and Excess factor is simply by having two different starting compounds but however you end with the same compound. That being said the compound with the most grams left over is the Excess reactant and the compound with the smaller amount of grams left over is the Limiting reactant.

For example if you have 5.25 g of Ca CO3 and 12.18 g of HNO3 and you wanted to know how much Carbon dioxide the two can hold you would go mass to mass on both to carbon dioxide then you determine which compound ends you with less grams of carbon dioxide. The answer is Ca CO3 which leaves you with 2.31 g of CO2.

## Theoretical Yield

The theoretical yield is the limiting reactant so for example 2.31 g of CO2 is the theoretical yield because it is the limiting reactant.

## Percent Yield

The percent yield formula is the actual yield/the theoretical yield * 100. The actual yield is given to us in this case which is 2.09 grams.

Solve: 2.09 g/2.31 g * 100 = 90.5%