HAPTER 3 - CHEMICAL REACTIONSStudyPage 1 of 3Next OReferencesUse the References to access important values if needed for this question.hen the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are:sulfur (s) +carbon monoxide (g)sulfur dioxide (g) +carbon (s)sp.cengage.info/als-asp/takea80°FO Type here to searchCa

Balanced equation :- A chemical equation is said to be balanced if number of atoms of different…

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Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter11: Solving Equilibrium Problems For Complex Systems

Section: Chapter Questions

Problem 11.3QAP

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Consider the generic chemical equationaA+bBcC+dD (where a, b, c, and d represent coefficients for the chemicals A, B, C, and D, respectively). a. How many possible values are there for “c”? Explain your answer. b. How many possible values are there for “c/d”? Explain your answer.
Please find the CO2 amount and show the calculation: Air composition (apprx) = 80% N2 and 20% O2 (mol basis only)Average molecular weight of Air = (0.8 x 28 + 0.2 x 32) = 29 g/mol (apprx) Math Sample:2kg pure charcoal (12C) is to be burnt completely with air. Find the air, CO2 andN2 amount in kg’s.Solution:C + O2 = CO2 (you must use a balanced equation)Therefore from the mole ratio of the reaction we write,C : O2 : CO2 = 1:1:1and from air composition we gotO2 : N2 : Air = 1:4:5 Given, 2 kg C = 2000 g/ (12 g/mol) = 166.67 mole CTherefore, similar mole of O2 required. So equivalent Air supposed to be 5times than the mole amount of O2 and released N2 will be 4 times than therequired O2. Therefore, Air amount = 5 x 166.67 moles= 833.34 moles= (833.33x29/1000) kg= 24.17 kg air Similarly N2 released amount will be = (166.67 x 4 x 28/1000) kg = 18.67 kgFind CO2 amount by yourself! (Isn’t it 7.34 kg?)
Assume that 1.00 mL (1.02 g) of crude product mixture is obtained from the reactionbefore the washing steps, and the distribution constant of the product mixture in brine isK = (Cmixture/Cbrine) = 50.0. Assuming that after mixing the product mixture volume isstill 1.00 mL (i.e., the volume lost is small), and the volume of the first brine wash layeris 2.00 mL. What mass (grams) of product mixture is lost (dissolves) in the first brinewash? (See Technique 13.2, p 53.) Show work.
Pure CO2 may be prepared by treating limestone with sulfuric acid. The limestone used in the process contains CaCO3, MgCO3 and inert. The acid used contains 12% H2SO4 by weight The residue from the process had the following composition. CaSO4 – 8.56%; MgSO4 – 5.23%; H2SO4 – 1.05%; Inert – 0.53%; CO2 – 0.12%; H2O – 84.51%; During the process, the mass was warmed where CO2 and H2O were removed. Calculate the following.a) The analysis of lime stone and b) The % excess acid used.
42. The primary use of 1,2‐dichloroethane, ClCH2CH2Cl, is to make vinyl chloride, which is then converted into polyvinyl chloride (PVC) for many purposes, including plastic pipes. Balance the following equation, which describes the industrial reaction for producing 1,2‐dichloroethane. C2H4 + HCl + O2 → ClCH2CH2Cl + H2O 50. Gaseous propane and liquid diethyl ether mix to form a solution. Which of thesesubstances is the solute and which is the solvent? 51. Consider a solution of 10% liquid acetone and 90% liquid chloroform. Which of these substances is the solute and which is the solvent?
Given Active Ingredient: precipitated sulfur (ointment) Raw Materials: 500 g calcium polysulphide and 1.5 kg hydrochloric acid Actual Yield: 343.4g precipitated sulfur Formulation: 250 mg per jar Dosage form: Ointment packaging:100 jars per box Synthesis and Packaging (Need answer)- Balanced Chemical Equation:- % composition by mass of each compound:- Mass to Mass Stoichiometry Calculation:- Limiting Reagent:- Excess Reagent:- Amount (g) in excess: % Yield:- Number of dosage form and packaging that can be produced from stoichiometric solution:
Give only typing answer with explanation and conclusion Balance the following equation 653 kJ + PCl3 ---> P + Cl2
In the common process for the manufacture of nitric acid, sodium nitrate is treated with aq. sulphuric acid containing 95% H2SO4. In order that the resulting "niter cake" may be fluid; it is desirable to use sufficient acid so that there will be 34% H2SO4 by wt in the final cake. This excess H2SO4 will actually be in combination with the Na2SO4 in the cake, forming NaHSO4, although for purposes of computation it may be assumed to be free acid. The cake will contain 1.5% water and the reaction will go to completion, but 2% of the HNO3 formed will remain in the cake. Assume that the NaNO3 used is dry and pure. Calculate per 100 kg of NaNO3 charged: (a) wt and % composition of the niter cake (b) wt of aqueous acid to be used (c) wt of water vapor and nitric acid distilled from the niter cake.
Predict the products for the following combustion reaction and write a balanced molecular equation. Be sure to include all phase labels. C3H80(l) + O2(g) ——-> ?
Using the provided data table, calculate ΔrxnH for the following reactions:a) 2O3(g) ⇌ 3O2(g)b) H2S(g) + 3/2 O2(g) ⇌ H2O(ℓ) + SO2(g) Compound ΔfH (kJ mol–1)H2O(ℓ) –285.83H2S (g) –20.6O2 (g) 0O3 (g) 142.7SO2 (g) –296.81
A 10 N aqueous sulfuric acid solution (300 Tw) is made to react with a 55% aqueous NaOH solution (640Be, heavier than water) to form 135 g of Na2SO4 per J 00 mL NaOH solution. The volume ratio of the acid solution to the base solution is 2: 1. Find: a) the Limiting reactant; excess reactant b) % excess reactant c) Degree or completion and d) % composition by weight of the materials after reaction.
The reaction C 7H 16 + O 2 ---> CO 2 + H 2O can be balanced with the coefficients

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Use the References to access important values if needed for this question. Then the following molecular equation is balanced using the smallest possible integer coefficients, the values of these coefficients are: sulfur (s) + carbon monoxide (g) sulfur dioxide (g) + carbon (s)