Chemistry 1050 Laboratory Fall 2016-2017 Experiment 6-Ga Constant NAMENathanPaun

Question

Chemistry 1050 Laboratory Fall 2016-2017 Experiment 6-Ga Constant NAMENathanPaune NAME Nathan Youre BENCH #-22 LAB SLOT LABORATORY REPORT DATA TABLES TABLE IAL 3 TRIAL TRIAL2 0.014 Mass of Mg () Volume 1.0 M HCI (mL Mass of flask (g) Mas of fask + HO 21670 |216709 Mass of flask + Ho Volume of flask (mL)12.25 m Initial presure aPa)l02.2o hl Initial pressure (kPa) 102.20 h lo2.I4hPa o2.20 nPa Final presure aPa) 135 he Final pressure (kPa) 47.0K Temperature (K) Instructor's initials and date: CALCULATIONS 1. Caleulate the moles of hydrogen gas produced from the reaction. (2 marks) Page 4 of 6

Explanation / Answer

Mg +2 HCl MgCl2 + H2

According to stoichiometry

24.3 g Mg reacts with 2 x 36.5 g HCl to give 2 g  H2 (1 mole hydrogen gas)

Here 10 ml 1M HCl means 0.01 mole HCl in 10 ml (1 x 10 /1000)

So in trial 1 we take 0.0129 g Mg

0.01 mole HCl = 0.365 g HCl

71 g HCl need 24.3 g Mg

Hence 0.365 g HCl needs 24.3 x 0.365/71 = 0.1249 g

so here limiting reagent is Mg since it present in small amount

0. 0129 g Mg = 0.0005 mole Mg

1 mole Mg produces 2 mole hydrogen

So 0.0005 mole produces 0.0005 mole x 2 = 0.001 mole H2

This is the easy way of calculating theoretically

Experimentally we can calculating this by collecting hydrogen gas over water.Change in pressure difference gives pressure of hydrogen at a particular temparature.

PV = nRT

Mass of flask = 74.45 g

Mass of flask + water = 216.70 g

Mass of water = 142.25 g

Volume of water = 142.25 ml

Volume of flask =

Empty volume=

Pressure difference P = 9.15 kPa = 0.090 atm

n = RT / PV =( 0.0821 x 298.1)/ (0.09 x empty volume)

Here data looks wrong.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.