Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

xe@chegg Study l Guided S Files × ? ? C Secure ! https://canvas.hawkeyecollege.e

ID: 1041412 • Letter: X

Question

xe@chegg Study l Guided S Files × ? ? C Secure ! https://canvas.hawkeyecollege.edu/courses/16860/files/folder/Take% HMy Hawkeye! Log in 001%20Exams?preview 964334 : Apps O canvas pOutlook.com-Micr Google Docs e chegg Study N Netflix a Amazon ? Google Calendar CHM-122-E3TH.docx Download load ?Info ×Close Page of 2 1. (5) Solid Fe reacts with aqueous sulfuric acid (H2SO4) to form aqueous iron (IIl) sulfate and hydrogen gas. Write out and balance the chemical equation, including phases. Dashbo 2. (5) What type of reaction is this? Can it undergo redox? If so, write out the balanced half reactions. 3. (10) You have 25.3 g of Fe and 65.8 g of H2S04; which reactant is limiting for the production of hydrogen gas, and how much hydrogen gas will this produce?

Explanation / Answer

1) 2 Fe(s) + 3 H2SO4(aq) --------------- Fe2(SO4)3 + 3 H2(g)

2)This is single displace ment reaction.

2 Fe(s) + 3 H2SO4(aq) --------------- Fe2(SO4)3 + 3 H2(g)

   0             +1                         +3                    0

So ,It is redox reaction

Fe undergoes oxdiation and H2 undergoes reduction.

oxidation half reation                                  Reduction half reaction

Fe ------------------- Fe+3                          H+ ----------------- H2

Fe ---------------- Fe+3 + 3e-                 2 H+ + 2e- -------------- H2

3)mass of Fe=25,3 grams

molar mass of Fe= 55.85 gram/mole

number of moles of Iron= 25.3/55.85 =0.453 moles

mass of H2SO4 = 65.8 grams

molar mass of H2SO4 = 98.0grams

number of moles of H2SO4 = 65.8/98=0.671 moles

according to eqaution

2 moles of Iron = 3 moles of H2SO4

0.453 moles of Iron= ?

                                  = 0.453x3/2= 0.6795 moles of H2SO4

number ofmoels of H2SO4 = 0.6795 moles

we need 0.6795 moles of H2SO4 .we have 0.671 moles of H2SO4. so it is limiting reagent

according to equation

3 moles of H2SO4 = 3 moles of H2

0.671 moles of H2SO4 = 0,671x3/3= 0.671 moles of H2

number of moles of H2 = 0.671 moles

molar mass of H2 = 2.0grams

mass of 0.671 moles = 2.0x0.671= 1.342 grams

mass of H2 produced= 1.342 grams.