Bellwork Monday A student drops a 3.40g piece of zinc

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Bellwork Monday A student drops a 3.40g piece of zinc into hydrochloric acid. How many liters of hydrogen gas are produced if the reaction occurs at STP?

Unit 8 Ch 12 Part 3 Limiting Reactants

12.3 Reactants in Excess and Limiting Reactants Limiting Reactant - the reactant that is used up in a reaction. When you run out of a reactant, the reaction stops and no more product is formed. Excess Reactant - the reactant that is not used up in a chemical reaction. The limiting reactant determines the amount of product formed. Important

Back to S’mores The s’mores can teach us one more lesson; the lesson of limiting reagents. Lets look at our balanced equation one more time. 3 chocolate squares + 1 marshmallow + 2 graham crackers  1 s’more If your friend is throwing a slumber party and has 137 chocolate squares and 47 marshmallows, how many total smores can be made. You will need to do the calculation with BOTH starting reactants. 137 choc.squares 1 s’more = 45.6 s’mores 3 choc.squares 47 marshmallows 1 s’more = 47 s’mores 1 marshmallow So, in this case, what LIMITS our production of s’mores? Chocolate squares! So chocolate squares are the limiting reagent!

Steps to Determining the Limiting Reactant Find out how much could be produced from each amount of reactant. Which ever produces less is limiting the amount produced. It is therefore the Limiting Reactant.

EXAMPLE If 6.70 mol of Na reacts with 3.20 mol Cl2, what is the limiting reagent? How many moles of NaCl will be produced? Which amount produced is less? 2 Na + Cl2 → NaCl 2 6.70 mol Na 2 mol NaCl = 6.70 mol NaCl 2 mol Na 3.20 mol Cl2 2 mol NaCl 6.40 mol NaCl = 1 mol Cl2 Cl2 is the limiting reactant This is the amount produced, and determines which is the limiting reactant

To determine how much excess reactant is left after a reaction, subtract how much of the excess reactant reacted from how much excess reactant you started with.

EXAMPLE When 0.50 mole of aluminum reacts with 0.72 mole of iodine, to form aluminum iodide, AlI3(s), what mass of AlI3 is produced? How much of the excess reactant will remain? I will separate this problem into two different steps. Find the limiting reactant and amount produced (as in previous example) Find amount in excess (left over)

When 0. 50 mole of aluminum reacts with 0 When 0.50 mole of aluminum reacts with 0.72 mole of iodine, to form aluminum iodide, AlI3(s), what mass of AlI3 is produced? Al + I2 → AlI3 2 3 2 0.50 mol Al 0.72 mol I2 2 mol AlI3 407.7 g AlI3 203.9 g AlI3 = 2 mol Al 1 mol AlI3 2 mol AlI3 407.7 g AlI3 = 195.7 g AlI3 3 mol l2 1 mol AlI3 amount produced I2 Limiting Reactant?

First…take the Limiting Reactant and the amount with which you began… How much of the excess reactant will remain? Al + I2 → AlI3 2 3 2 First…take the Limiting Reactant and the amount with which you began… …then convert to find the amount that reacted with it… 0.72 mol I2 2 mol Al 0.48 mol Al = 3 mol l2 …then subtract the amount used from the amount given (starting amount) 0.50 mol Al - 0.48 mol Al = 0.02 mol Al given in the problem this is the excess (left over)

Bellwork Tuesday 5.9 L of carbon dioxide is combined with 8.4 g MgO in a synthesis reaction to form magnesium carbonate. How many grams of magnesium carbonate is created in this reaction?