Chemistry 1: PROBLEM SET SY 2012/2013 CLASS #: _______________ NAME_________________________________________ SECTION: ______________________ Stoichiometry II - Mole Calculations/ Limiting and Excess Reagent – Lecture Notes 1. Given the balanced equation N2(g) + 3H2(g) 2NH3(g) How many moles of ammonia are produced when 0.60 mol of nitrogen reacts with hydrogen? 2. Given the equation: SiO2 + HF SiF4 + H2O a. Calculate the number of moles HF that would completely react with 2.5 moles of
Premium Chlorine Sodium Stoichiometry
Stoichiometry lab By: James Stewart Purpose: To calculate mole ratios Introduction: There are two types of chemical analysis; qualitative analysis which is the identification of a substance present in a material‚ and qualitative analysis which measures the amount of the substance. In this lab‚ you will perform a quantitative analysis of a two-step reaction. Copper(II) oxide will be synthesized from a known mass of copper(II) sulfate pentahydrate. Using the relationship of the balanced equation
Premium Stoichiometry Yield Sodium hydroxide
combining the elements‚ instead of having it naturally do so‚ made a significantly fully more bright color. In Theophilus’s “recipe” for vermillion‚ it is stated that there should be far more sulfur than a stiochiometric reaction should require. Stoichiometry
Premium Chemistry Chemical reaction Oxygen
STOICHIOMETRY OF GASOLINE. AN INTRODUCTION The internal combustion engines burn fuel to create kinetic energy. The burning of fuel is basically the reaction of fuel with oxygen in the air to form water and Carbon dioxide as the major end product . The amount of oxygen present in the cylinder is the limiting factor for the amount of fuel that can be burnt that is to say it determines the level of burning in our combustion engine. If there’s too much fuel present‚ not all fuel will be burnt and un-burnt
Premium Internal combustion engine Combustion Gasoline
moles-stoichiometry-practice-problems Now you’re ready to use what you know about conversion factors to solve some stoichiometric problems in chemistry. Almost all stoichiometric problems can be solved in just four simple steps: 1.Balance the equation. 2.Convert units of a given substance to moles. 3.Using the mole ratio‚ calculate the moles of substance yielded by the reaction. 4.Convert moles of wanted substance to desired units. These "simple" steps probably look complicated at first
Premium Mole Atom Molecule
Name: Brian James|Date:3/10/13| Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: 73426| Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3(s)= CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =.68 g Step 4: Mass of weighing dish _0.6_g Mass of weighing dish and Na2CO3 .72_g Net mass of the Na2CO3 .12_g Step 6: Mass of filter paper
Premium Stoichiometry Sodium bicarbonate Yield
Stoichiometry I. Introduction/ Purpose: Stoichiometry is the study of the quantitative‚ or measurable‚ relationships that exist in chemical formulas and also chemical reactions. The calculations of a stoichiometry problem depend upon balanced chemical equations. The coefficients of the balanced equations indicate the molar ratio of the reactants and products taking part in the reaction. There are three major categories of stoichiometry problems such as mass-mass‚ mass-volume‚ and volume-volume
Premium Stoichiometry Chemical reaction
.6Name: Danielle Hall|Date: 10.15.2012| Exp 9: Stoichiometry of a Precipitation Reaction|Lab Section: | Data Tables: Step 3: Show the calculation of the needed amount of Na2CO3 CaCl2.H2O(aq)= m/M =1/147 =0.0068 mol CaCO3(s)=0.0068*1/1 =0.0068 mol CaCO3(s)= CaCO3 (s)= CaCO3 mol *CaCO3 g =0.0068 mol*100.01 g =.68 g Step 4: Mass of weighing dish _0.6___g Mass of weighing dish and Na2CO3 __0.72__g Net mass of the Na2CO3 ___0.12_g Step 6: Mass of filter paper
Premium Stoichiometry Solubility Measurement
Chem 121L Part I: Introduction Stoichiometry is the study of the quantitative‚ or measurable‚ relationships that exist in chemical formulas and also chemical reactions. In this experiment hydrogen gas will be produced from the reaction of a known mass of magnesium metal with an excess of hydrochloric acid. The theoretical number of moles of hydrogen gas may be calculated using stoichiometry and the balanced chemical equation. The theoretical volume of hydrogen gas may then be determined from
Premium Chemical reaction Gas Ideal gas law
Chapter 4 — Intro—1 1 CHAPTER 3 Topic Scopes: Stoichiometry and Solution Concentration • Molarity‚ molality‚ parts per million & percentage (w/w‚ w/v and v/v) • Stoichiometry calculation • Limiting reactant • Theoretical yield‚ actual yield and percentage yield 1 2 Mole Concept No. of Moles = Molarity (M) • Molarity (molar concentration) is the number of moles of a solute that is contained in 1 liter of solution Mass (g) molar mass (g/mol) No. of Moles = Molarity (mol/L) volume (L) Molarity
Premium Stoichiometry Concentration Mole