Chemistry I Chapter 9 Review
Actual yield = amount of product that is
recovered
Excess reactant = reactant that remains when
reaction ends
Limiting reactant = reactant that runs out first
Percent yield = comparison of predicted and
actual amounts of product
Chemistry I Chapter 9 Review
Stoichiometry = calculation of quantities in
chemical reactions
Theoretical yield = predicted amount of product
*
Mass and atoms are conserved in a chemical
reaction. (why we have to balance equations!)
Chemistry I Chapter 9 Review
Actual yield = amount from experiment
Theoretical = CALCULATED AMOUNT!!!!
Mole ratio (mol over mol) – use the coefficients
Always convert the first component from grams
to moles.
Balance - Al + CuSO4  Al2(SO4)3 + Cu
Interpret the equation for atoms and moles.
Chemistry I Chapter 9 Review
If 0.5moles of Al(NO3)3 are used up in the
following reaction, how many moles of HNO3
will be produced?
Al(NO3)3 + H2SO4  Al2(SO4)3 + HNO3
Balance first then solve for 0.5mol.
Chemistry I Chapter 9 Review
If 50g of iron react with oxygen gas to produce
iron(III)oxide, how many moles of oxygen gas
will be needed?
1-write equation
2-balance equation
3-change 50 to moles
4-use mole ratio
Chemistry I Chapter 9 Review
Zinc sulfide reacts with oxygen gas to form zinc
oxide and sulfur dioxide. If 2moles of ZnS are
combined with 3 moles of O2, how many moles of
ZnO can be produced? How many moles of the
excess reactant will be left over?
1- write and balance equation
2- do each reactant separately
3- the lesser mole amount “wins” (is the answer)
4-to find how much excess, calc and sub from
original starting moles.
Chemistry I Chapter 9 Review
Solid iron reacts with hydrochloric acid to form
iron(II)chloride and hydrogen gas.
What mass of FeCl2 could be produced from 70g
of Fe and excess of HCl?
1-write and balance equation
2-change 70g to moles
3-use mole/mole ratio (and coefficients)
4-change moles back to grams
Chemistry I Chapter 9 Review
Magnesium carbonate decomposes into
magnesium oxide and carbon dioxide. If 23.4g of
MgO is produced from 59.6g of MgCO3, what is the
percent yield?
1-write and balance equation
2-change g to mole
3-use mole / mole ratio
4-change moles back to grams
5-%yield = actual divided by theoretical times 100