Tutorial 3 THE MOLE AND STOICHIOMETRY
Tutorial 3 THE MOLE AND STOICHIOMETRY A chemical equation shows the reactants (left side) and products (right side) in a chemical reaction. A balanced equation shows, in terms of moles, how much of each substance is involved in the reaction. Stoichiometry is the study of the relationships of quantities of substances in a chemical reaction.
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Clark, Smith (CC-BY-4.0) GCC CHM 130 Chapter 13: Stoichiometry page 3 13.4 Volume-Volume Stoichiometry Molar Volume gas @ STP Fact: If you start with liters of the given and are asked to find liters of the unknown, as long as the gases are at the same temperature and pressure the molar volumes will cancel out with each other so you are ...
Gas Stoichiometry . Chemistry 110. 1] Given the equation: 2 NH3 (g) + 3 Cl 2(g) ---> N (g) + 6 HCl(g) a. How many milliliters of nitrogen can be made from 13 L of chlorine and 10.0 L of ammonia gas at STP? 10.0 L NH3 X . 1 L N. 2. 2 L NH3 = 5.00 L N. 2.
Answers: Moles and Stoichiometry Practice Problems 1) How many moles of sodium atoms correspond to 1.56x1021 atoms of sodium? 1.56 -x 1021 atoms Na x 1 mol Na = 2.59 x 10 3 mol Na 236.022 x 10 atoms Na 2) Determine the mass in grams of each of the following: a. 1.35 mol of Fe 1.35 mol Fe x 55.845 g Fe = 75.4 g Fe
Equation Stoichiometry . Chemistry 110 . 1] Given the equation: 2C8H18 + 25 O2----> 16CO2 + 18H2O . a. How many moles of oxygen gas are required to make 8.33 moles of carbon dioxide? ... How many moles of Al4C3 were reacted when 3.55 x 10 35 formulas units of aluminum hydroxide were produced? 3.55 x 1035 formula units Al(OH)3 X . 1mol Al(OH)3 6 ...
STOICHIOMETRY AND PERCENT PURITY Many samples of chemicals are not pure. We can define percent purity as mass of pure compound in the impure sample total mass of impure sample ... 2 needed to make 127 g of Mg3(PO4)2, assuming a 100% yield. mass Mg(OH)2 = 127 g Mg3(PO4)2 x 1 mole Mg3(PO4)2 x 58.3 g Mg(OH)2 1 mole Mg(OH)2 1 mole Mg3(PO4) 3 …
Reaction Stoichiometry Moles of The coefficients in a balanced chemical equation specify the relative amounts in moles of each of the substances involved in the reaction. 2 C8H18 (l) + 25 O2(g) 16 CO2 (g) + 18 H2O(g) 2 mol C8H18: 25 mol O2: 16 mol CO2: 18 mol H2O How many grams of glucose can be synthesized from 37.8 g of CO2 in photosynthesis?
3 PO 4 x 163.94 g Na 3 PO 4 = 126.75 g FeCl 2 3 mole FeCl 2 1 mole Na 3 PO 4 = 20. g Na 3 PO 4 Since we have 41 g Na 3 PO 4, FeCl 2 is the limiting reagent. 23 g FeCl 2 x 1 mole FeCl 2 x 6 mole NaCl x 58.44 g NaCl = 126.75 g FeCl 2 3 mole FeCl 2 1 mole NaCl = 21 g NaCl 3) How much of the excess reagent remains when this reaction has gone to ...