{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 6 2h2 o2 2h2o h 571 co c o2 h 1105 co2 h 3935 c

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: .67 Moles of HNO3 available 84/63 = 1.333 Stoichiometry ratio must be 3/8 HNO3 is the limiting reagent gr Cu(NO3)2 = 187.5 x moles gr = 93.8 moles = moles HNO3 /8 x 3 = 1.333/8 x 3 5. 6. - 4OH + Br 2 + M n 2+ ----- MnO2 + 2Br - 12 + 2H2O total coeff = 11 moles H2SO4 = moles NaOH / 2 2 V x 0.110 = 10 x 0.085 7. - V = 3.9 mL moles HNO3 = 23.4 x 0.601 / 1000 = 0.0140 Moles Cu = moles HNO3 x 3/8 = 0.00527 gr Cu = 0.00527 x 63.5 = 0.334 8. 81.4% moles gas = PV / RT = 0.5 Mw = gr /moles = 64.2 SO2 is the only possibility 9. Mole He = 1.37 x 2.05/760 / 0.082 / 297.5 = = 0.0151 Moles Ne = 721/1000 x 0.185 / 0.082 / 309.2 = 0.00526 Total mole = moleHe + mole Ne = PV/RT from which P = 0.253 atm 10. Using PV = nRT calculate the V at the initial conditions = 8.11 L Work = V DP x 101.3 (conversion factor for joules) = -2.73 103 11. CO2 + 2H2O -CH3OH + 3/2 O2 ΔH° = +726.6 2H2 + O2 -2H2O ΔH° = -571 CO -C + ½ O2 ΔH° = +110.5 ---CO2 ΔH° = -393.5 C + O2 __________________________________________________________________ 2H2 + CO 12. C8H18 ----- +...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern