Problem 8.53 Part A Mg metal reacts with HCl to produce hydrogen gas Mg(s) +2HCl

Question

Problem 8.53 Part A Mg metal reacts with HCl to produce hydrogen gas Mg(s) +2HCl(aq)-MgCl2 (aq) H2 (g) What volume of hydrogen at 0 C and 1.00 atm (STP) is released when 8.15 g of Mg reacts? Express your answer with the appropriate units. v- 22.4 Submit My Answers Give Up Incorrect; Try Again; 3 attempts remaining Part B How many grams of magnesium are needed to prepare 7.70 L of H2 at 730 mmHg and 23 C? Express your answer with the appropriate units. LA m= 1 Value Units Submit My Answers Give Up

Explanation / Answer

part A

   1 mol Mg = 1 mol H2

no of mol of Mg = 8.15/24 = 0.34 mol

no of mol of H2 = 0.34 mol

volume of H2 = nRT/P

               = 0.34*0.0821*273.15/1

               = 7.625 L

part B

volume of H2 = nRT/P

             7.7 = n*0.0821*296.15/(730/760)

   n = no of mol of H2 = 0.304 mol

mass of Mg REQUIRED = 0.304*24 = 7.296 grams

part A

volume of gas = nRT/P

               = 4.1*0.0821*(273.15-13)/(0.8)

               = 109.46 L

part B

volume of gas = nRT/P

           18.0 = 5*0.0821*(273.15+9)/p

p = pressure of gas = 6.434 atm

part C

volume of gas = nRT/P

           300 = n*0.0821*(273.15+1285)/1.025

n = no of mol of gas = 2.404 mol

part D

volume of gas = nRT/P

           18 = 7.6*0.0821*T/3.6

T = temperature = 103.85 k

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