# Stoichiometry

### with baking soda and hydrochloric acid

## Balancing the equation

First and foremost you have to balance your equation. Today we have the baking soda NaH(CO3) and hydrochloric acid (HCl) they yield NaCl + H2O + CO2. If you noticed, we have three products and that's because of decomposition. When you look back at the equation you need to make sure the there is the same number of each element is on each side.

NaH(CO3) + HCl ---> NaCl + H2O + CO2

Everything is the same on each side, so that means that all they're coefficients is 1.

NaH(CO3) + HCl ---> NaCl + H2O + CO2

Everything is the same on each side, so that means that all they're coefficients is 1.

How to Balance Chemical Equations & Reactions 1 - EASY!

## Mole to Mole Conversions

Next we have mole to mole conversions. You use mole in this not grams! Grams will come later. You find the amount of moles for your element or compund by looking at their coefficients in your balanced equation. Here we're trying the amount of NaCl we'll get out of 11.30g NaH(CO3).

You put your given on the top left diagonally from it you put the number of moles (from your balanced equation) of NaH(CO3), and above that you put the number of moles of NaCl. Now multuply across the top and divide that by the bottom.

11.30mol NaH(CO3)| 1mol NaCl

------------------------------------------------------- = 11.30 mol NaCl

|1mol NaH(CO3)

You put your given on the top left diagonally from it you put the number of moles (from your balanced equation) of NaH(CO3), and above that you put the number of moles of NaCl. Now multuply across the top and divide that by the bottom.

11.30mol NaH(CO3)| 1mol NaCl

------------------------------------------------------- = 11.30 mol NaCl

|1mol NaH(CO3)

Stoichiometry - Moles to Moles (using a balanced equation) | www.whitwellhigh.com

## Mass to mass

We start with what we know which is 10.03g of NaH(CO3) which will go on the top left corner. Diagonally from it we put the molar mass of NaH(CO3)which we will get from the periodic table, Above that we will put 1mol of NaH(CO3) because you will always put 1mol of element/compound A there. diagonally from that we put the the number of moles of NaH(CO3) (coefficients from balanced equation). Above that we put the number of moles we for NaCl which is our first product. Diagonally from that we put will forever put 1mol of element/compound B. Multiply across the top, multiply across the bottom, them divide the bottom by the top.

10.03g NaH(CO3) | 1mol NaH(CO3) | 1mol NaCl |58.443g NaCl

--------------------------------------------------------------------------------------------------- = 6.98g NaCl

| 84.006g NaH(CO3)|1mol NaH(CO)|1mol NaCl

10.03g NaH(CO3) | 1mol NaH(CO3) | 1mol NaCl |58.443g NaCl

--------------------------------------------------------------------------------------------------- = 6.98g NaCl

| 84.006g NaH(CO3)|1mol NaH(CO)|1mol NaCl

Stoichiometry: Mass-Mass

## Limiting and excess reactant

Limiting and excess is done to find which element/compound will be the limit of how much you can make, and you find out the excess you have left. This will be set up like mole to mole.

We start with what we know which is 5.25g of NaH(CO3) which will go on the top left corner. Diagonally from it we put the molar mass of NaH(CO3)which we will get from the periodic table, Above that we will put 1mol of NaH(CO3) because you will always put 1mol of element/compound A there. diagonally from that we put the the number of moles of NaH(CO3) (coefficients from balanced equation). Above that we put the number of moles we for CO2 which is our first product. Diagonally from that we put will forever put 1mol of element/compound B. Multiply across the top, multiply across the bottom, them divide the bottom by the top.

You will do the same for the second equation, but in stead of using NaH(CO3) as compound A, we'll be using HCl.

5.25g NaH(CO3) | 1mol NaH(CO) | 1mol NaCl |44.009g CO2

--------------------------------------------------------------------------------------------------- = 2.75g CO2

|84.006g NaH(CO3) |1mol NaH(CO)|1mol NaCl

12.18g HCl | 1mol HCl | 1mol CO2 | 44,009g CO2

----------------------------------------------------------------------------- = 14.70g CO2

|36.461g HCl| 1mol HCl |1mol CO2

The lowest outcome out of the two will determine you limit reactant. NaH(CO3) is the limiting reactant. HCl is the excess since it had a bigger product of CO2.

We start with what we know which is 5.25g of NaH(CO3) which will go on the top left corner. Diagonally from it we put the molar mass of NaH(CO3)which we will get from the periodic table, Above that we will put 1mol of NaH(CO3) because you will always put 1mol of element/compound A there. diagonally from that we put the the number of moles of NaH(CO3) (coefficients from balanced equation). Above that we put the number of moles we for CO2 which is our first product. Diagonally from that we put will forever put 1mol of element/compound B. Multiply across the top, multiply across the bottom, them divide the bottom by the top.

You will do the same for the second equation, but in stead of using NaH(CO3) as compound A, we'll be using HCl.

5.25g NaH(CO3) | 1mol NaH(CO) | 1mol NaCl |44.009g CO2

--------------------------------------------------------------------------------------------------- = 2.75g CO2

|84.006g NaH(CO3) |1mol NaH(CO)|1mol NaCl

12.18g HCl | 1mol HCl | 1mol CO2 | 44,009g CO2

----------------------------------------------------------------------------- = 14.70g CO2

|36.461g HCl| 1mol HCl |1mol CO2

The lowest outcome out of the two will determine you limit reactant. NaH(CO3) is the limiting reactant. HCl is the excess since it had a bigger product of CO2.

STOICHIOMETRY - Limiting Reactant & Excess Reactant Stoichiometry & Moles

## Percent yield

Percent yield you get by putting your given over the theoretical yield and times that by 100.

2.60

------- x 100 = 94.5%

2.75

2.60

------- x 100 = 94.5%

2.75

## Theorectical yield

The theoretical yield is actually the number you got for you limiting reactant. So our theoretical yield is 2.75 CO2

2.75g CO2

2.75g CO2

Calculating Theoretical & % Yield

Baking Soda and Hydrochloric Acid