When hydrochloric acid (HCI) reacts with sodium sulfide (Na,S), hydrogen sulfide (H,S) gas andsodium chloride (NaCl) are produced. How many moles of H,S will be generated by 2.56 molesof HCl? Show work!!!2HCI + Na,S->H,S + 2NACI.1.28 moles2.56 moles5.12 moles5.56 moles(Ratio per equation)(moles produced if) How many moles of H,0 (water) are produced when 3.25 moles of CO, (Carbon Dioxide) are formerduring the combustion of CH,OH (methyl alcohol)? Show work2CH;OH + 30,>2CO,+ 4H,0(Ratio per equation)(moles produced if)1:2 moles of CO,3.25moles of CO2X moles of H,03.25 moles of H,01.62 moles3.25 moles6.50 moles13.00 molesAa..

Answered: Image /qna-images/answer/977e1e2d-86f9-4874-90d8-e3b40b1c5bff.jpg

When hydrochloric acid (HCI) reacts with sodium sulfide (Na,S), hydrogen sulfide (H,S) gas and sodium chloride (NaCl) are produced. How many moles of H,S will be generated by 2.56 moles of HCI? Show work!!! 2HCI + Na,s->H,S + 2NACI. 1.28 moles 2.56 moles 5.12 moles 5.56 moles (Ratio per equation) (moles produced if) %3D

When hydrochloric acid (HCI) reacts with sodium sulfide (Na,S), hydrogen sulfide (H,S) gas and sodium chloride (NaCl) are produced. How many moles of H,S will be generated by 2.56 moles of HCI? Show work!!! 2HCI + Na,s->H,S + 2NACI. 1.28 moles 2.56 moles 5.12 moles 5.56 moles (Ratio per equation) (moles produced if) %3D

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

how to calculate moles of NaHCO3 and moles of CO2 from Mass of NaHCO3= 0.081g Volume of CO2=5ml show steps 10mins
The hydrated salt, MY3·xH2O, was heated strongly in a crucible to remove all the watermolecules of crystallization.The data are for this experiment are given in the table below.Mass of empty crucible and cover 29.73 gMass of crucible, cover, and hydrated salt 32.72 gMass of crucible, cover, and anhydrous salt 31.52 gMolar mass of anhydrous salt 241.88 g/mol(i) Calculate the % composition by mass of water in the hydrated salt.(ii) Determine the value of x.(iii) M is in Period 4 of the Periodic Table of the Elements.If Y = NO3–, identify the element M.(iv) What is the name of this hydrated salt? ___
For the following table fill in the missing values and calculate the theoretical yield. A + 2B à C moles 0.00242 0.00242 ?_________ MW ?_________ 98.06 164.16 g ?_________ ?__0.00002467_______ ?_________ mL 0.20 d(g/mL) 0.80
What is the percent of each halogen in a 0.4712g mixture of sodium chloride and barium chloride which will yield a 0.9274g dried silver chloride. The MW for sodium chloride, barium chloride and silver chloride are 58.45g/mol, 208.25g/mol and 143.33 g/mol respectively.
A chemist carries out the reaction below and obtains 3.45g of iron Fe (Fe, MW=55.845g/mol). Fill in the blanks below to determine the percent yield of the reaction when 8.30g of Fe2O3 (MW= 159.69g/mol) and 3.50g of CO (MW=28.01G/mol)were used in the reaction. Fe2O3(s) + 3CO(g) = 2Fe(s) + 3CO2(g) Note that for two of the blanks, will be the same answer. 1. What would be the mass in grams of Fe produced based on the mass of Fe2O3 used? answer: (blank) 2. What would be the mass in grams of Fe produced based on the mass of CO used? answer: (blank) 3. What is the theoretical yield of Fe in grams? answer: (blank) 4. What is the limiting reagent? Enter the formula of compound. answer: (blank) 5. What is the percent yield of this reaction? answer: (blank)
On strong heating ammonia dissociates into nitrogen and hydrogen, all in a gaseous state. When 0.350 mol of NH3 is heated in a closed 10.0 L vessel, the fraction of dissociation is 0.400. (ensure all calculation have 3 decimal places. a. Calculate moles of NH3dissociated? b. Calculate moles N2 formed? c. Calculate moles H2 Formed?
For every one molecule of NaHCO3(s) there is one molecule of CO2(g) that forms: NaHCO3(s) + CH3CO2H(aq) --> CO2(g) + H2O(l) + NaCH3CO2(aq) This means that if we know how many moles of gas we need, we need the same number of moles of sodium bicarbonate to produce it. Now, we need something to fill with gas. I'll be filling a 1-quart sealable plastic bag. 1 quart is 946 mL (0.946 L). We can use the ideal gas law to determine how much gas (in moles) is needed to fill 946 mL. Remember there are four variables for the ideal gas law. Ideal Gas Law: PV = nRT P = pressure: typically 715 mm Hg in Spokane <-- unit dictates which R to use V = volume: 1 quart bag = 0.946 L <-- must be in Liters T = room temperature: 20 °C in my lab, +273 = 293 K <-- must be Kelvin n = moles: what we are solving for I used atmospheric pressure in mm Hg, so I'll have to pick the R value with "mm Hg" in the units: R = 62.4 L × m m H g K × m o l Now, use these values to calculate how many moles of…
Citric acid, C6H8O7, a component of jams, jellies, and fruity soft drinks, is prepared industrially via fermentation of sucrose by the mold Aspergillus niger. The equation representing this reaction is C12H22O11 + H2O + 3O2 --> 2C6H8O7 + 4H2O Using a metric ton(1000kg) of sucrose the expected yield is 1122Kg. On average if only 1036kg is produced, what is the % yield for this reaction?
C9H8O4 is synthesized by reacting stoichiometric amounts of C7H6O3 and (CH3CO)2O, as shown by the balanced chemical equation in the picture: During the synthesis, 500 grams of salicylic acid was available for use in the experiment. Other information: MW of Salicylic Acid: 138.121 g/mol MW of Acetic Anhydride: 102.09 g/mol Density of Acetic Anhydride: 1/08 g/ml Question: The available Acetic anhydride reagent in the lab is in the liquid form. How many mL of Acetic anhydride is needed to react with the available salicylic acid stoichiometirically?
2N2(g) + 3O2(g) ® 2N2O3(g) DH°rxn = +86.6 kJ 1. When 12.0 moles of N2 react, the reaction heat is ____________ kJ 2. To make +552 kJ of reaction heat, _________ moles of O2 must react. 3. To make +24 kJ of reaction heat, a count of ________________ N2O3 molecules must form.
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:
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?)