Daily Science pg. 82 You are doing the following reaction in the lab. You start with g of magnesium. How many grams of oxygen gas do you need? How much magnesium oxide will you make? 2Mg + O 2  2MgO

Limiting Reagents pg. 83

why reactions stop one reactant is in excess and one is limiting the reaction from proceeding.

Determining the limiting reagent The masses of all reactants are given must convert grams of each reactant into moles make a ratio of the moles of each divide the bigger number by the smaller compare the calculated ratio to the ratio of moles from the balanced equation If the numerator is less than the ratio from the reaction, it is the limiting reagent; if the numerator is greater than the numerator in the ratio, it is in excess

Calculating the amount of product start with the moles of the limiting reactant and convert to moles of the product by using mole to mole ratios convert from moles of the product to mass of the product by using molar mass

Determining the amount of excess Start with the moles of the limiting reactant convert to moles of excess using mole to mole ratios Convert to grams using molar mass Subtract the calculated mass from the given mass

Practice g of S 8 reacts with g of Cl 2. Which is the limiting reagent? Determine the amount of product that will be formed and how much excess reactant there is. S 8 + 4Cl 2  4S 2 Cl 2

Your turn Determine the mass of tetraphosphorus decaoxide formed if 25.0 g of phosphorus and 50.0 g of oxygen are combined. How much of the excess reagent remains? P 4 + 5O 2  P 4 O 10