Stoich Guide for Chemistry Dummies
By Joseph Warren
Double replacement 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/moleBa(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
(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
((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
((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
Percent Yield
100(2.35/2.75)=85.5%