Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Formulas, Equations and Moles - General College Chemistry I - Worksheet | CHEM 112, Assignments of Chemistry

Material Type: Assignment; Class: General College Chemistry I; Subject: Chemistry; University: University of Illinois - Chicago; Term: Fall 2002;

Typology: Assignments

Pre 2010

Uploaded on 07/29/2009

koofers-user-yxk-1
koofers-user-yxk-1 🇺🇸

10 documents

1 / 1

Toggle sidebar

Related documents


Partial preview of the text

Download Formulas, Equations and Moles - General College Chemistry I - Worksheet | CHEM 112 and more Assignments Chemistry in PDF only on Docsity! Chemistry 112 Worksheet — Formulas, Equations, and Moles Harwood KEY 1. The following diagram shows A (closed circles) reacting with B (open circles). Write a balanced chemical equation describes the stoichiometry of the reaction depicted in the diagram A2 + B2 → 2AB 2. How many moles does each of the following samples contain? a. 1.0 g of lithium (1.0 g)(1 mol / 6.941 g) = 0.14 mol b. 1.0 g of nitrogen gas (nitrogen gas exists as N2) (1.0 g)(1 mol / 28.0134 g) = 0.036 mol c. 1.0 g of penicillin G potassium, C16H17N2O4SK (1.0 g)(1 mol / 372.4856 g) = 0.0027 mol 3. What is the empirical formula of stannous fluoride, a compound added to toothpaste to protect teeth against decay? Its mass percent composition is 24.25% F, 75.75% Sn. mol F = (24.25 g)(1 mol / 18.9984 g) = 1.27642 mol F mol Sn = (75.75 g)(1 mol / 118.710 g0 = 0.638101 mol Sn F: 1.27642 mol / 0.638101 mol = 2 Sn: 0.638101 mol / 0.638101 mol = 1 Empirical formula = SnF2 4. Disilane, Si2Hx , is analyzed and found to contain 90.28% by mass silicon. What is the value of x? Mass of Si = 90.28 g per 100 g sample Mass of H = 100.00 g - 90.28 g = 9.72 g mol Si = (90.28 g)(1 mol / 28.0855 g) = 3.21447 mol mol H = (9.72 g) (1 mol / 1.0079 g) = 9.643814 mol Si: 3.21447 mol / 3.21447 mol = 1 H: 9.643814 mol / 3.21447 mol = 3 x = 3 5. How many grams of the dry-cleaning solvent ethylene chloride, C2H4Cl2, can be prepared by reaction of 15.4 g of ethylene, C2H4, with 3.74 g of Cl2? C2H4 + Cl2 → C2H4Cl2 mol ethylene = (15.4 g)(1 mol / 28.0536 g) = 0.5489 mol mol chlorine = (3.74 g)(1 mol / 70.906 g) = 0.05274 mol maximum mol ethylene chloride formed from all ethylene: mol = (0.5489 mol C2H4)(1 mol C2H4Cl2 / 1 mol C2H4) = 0.5489 mol C2H4Cl2 mol = (0.05274 mol Cl2)(1 mol C2H4Cl2 / 1 mol Cl2) = 0.05274 mol C2H4Cl2 So 0.05274 mol C2H4Cl2 will be formed. mass C2H4Cl2 = (0.05274 mol)(98.9596 g / 1 mol) = 5.22 g C2H4Cl2
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved