# Stoich Guide for Chemistry Dummies

## Double replacement reaction

Zinc nitrate mixed with barium hydroxide produces a solid zinc hydroxide and an aqueous Barium Nitrate. The zinc hydroxide is a solid due to the Solubility of Common Ionic Compounds chart.
This type of a reaction is called double replacement, because the two metals from aqueous solutions trade positions trade position during the reaction

## Zinc Nitrate and Barium Hydroxide

Zinc2+ nitrate- plus barium+2 hydroxide- yield zinc+2 hydroxide- with Barium+2 Nitrate-.

Zn(NO3)2(aq) + Ba(OH)2(aq) --> Zn(OH)2(s)+Ba(NO3)2(aq)

This is the balanced equation showing the charges the previous step. The charges cross to show how many of each element/polyatomic will make the compound. The first step would also be the IUPAC form of the equation. The IUPAC form repeats the equation but with words and not symbols.

## Molar Masses

Molar Mass is the mass of a mole of a single element or compound. The mass of each element can be found on the periodic table, so you just add the masses of all the elements within the compound to find the mass of the whole compound.

Zn(NO3)2=65.38+2(14.007)+6(15.999)=189.388g/mole

Ba(OH)2=137.328+2(15.999)+2(1.008)=171.342 g/mole

Zn(OH)2=65.38+2(15.999)+2(1.008)=99.394 g/mole

Ba(NO3)2=137.328+2(14.007)+6(15.999)=261.336 g/mole

## Mole to mole

Mole to mole is the conversion of moles of one compound to moles of another compound within a balanced equation. This uses the ratio of coefficients between compounds. In this reaction, a single mole of any compound will make a mole of every other. Most reactions are not this simple and might take 5 moles of O2 to make 10 moles of H20.

(11.30 mol Zn(NO3)2 * 1 mol Zn(OH)2)/(1mol Zn(NO3))=(11.30 * 1 mol Zn(OH)2)/1=11.30 mol Zn(OH)2

(12.1 mol Ba(OH)2 * 1mol Ba(NO3)2)/(1 mol Ba(OH)2)=(12.1 * 1 mol Ba(NO3)2)/1=12.1 mol Ba(NO3)2

## Mass to mass

Mass to mass is the conversion of grams to moles, then to moles of another compound using mole to mole, and mole to grams. This uses the molar mass that we found in the molar mass section.

((7.24g Zn(NO3)2 * 1 mol Ba(OH)2 * 1 mol Zn(OH)2*99.394g Zn(OH)2)/(171.342g Ba(OH)2 * 1 mol Ba(OH)2 * 1 mol ZN(OH)2))=4.20g Zn(OH)2

## Limiting and excess reactants

Reactions do not occur with perfects amounts of each compound to convert everything to another compound. There will be some reactant left over for most reactions. The reactant that stops the reaction from making more product is called the limiting reactant. To find the limiting reactant, you find the mass to mass conversion of each reactant to the same product. This reaction had 5.25g Zn(NO3) and 12.18g Ba(OH)2.

((5.25g Zn(NO3)2 * 1 mol Zn(NO3)2 * 1 mol Zn(NO3)2 * 99.394g Zn(OH)2)/(189.398g (NO3)2 * 1 mol Zn(NO3)2 * 1 mol Zn(OH)2))=2.75g Zn(OH)2

((12.18g Ba(OH)2 * 1 mol Ba(OH)2 * 1 mol Ba(OH)2 * 99.394g Zn(OH)2)/(171.342g Ba(OH)2 * 1 mol Ba(OH)2 * 1 mol Zn(OH)2))=7.066g Zn(OH)2

Since we made less zinc hydroxide with our zinc nitrate, this is our limiting reactant.

## Theoretical yield

Theoretical yield is the amount of product we would make in a perfect reaction with our limiting reactant. Since we are able to make a maximum of 2.75g Zn(OH)2, this is our theoretical yield.

## Percent Yield

Percent yield is the percentage of the amount you actually got in the real world after doing the experiment, divided by the amount you should have gotten, the theoretical yield. For this reaction, we had 2.35g as the product.

100(2.35/2.75)=85.5%