0.015.grams magnesium was placed in a 153 ml flask. The flask was immersed in a
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Question
0.015.grams magnesium was placed in a 153 ml flask. The flask was immersed in a water bath whose temperature was 24.0 degree C. When 5 mLs of 1.0 M HCI were injected into the flask the attached pressure sensor rose from 100.3 kPa to 108.7 kPa. Write the balanced chemical equation for magnesium reacting with HCI. What was the volume in mL of H_2 gas produced from the reaction? What was the temperature, in kelvin, of the H_2 gas? What was the pressure, in kPa, of H_2 gas? Using the combined gas law, calculate the volume of H_2 gas at STP. Calculate the number of moles of H_2 produced when all the magnesium reacts with the HCI. Calculate the molar volume, in units of liters/mole, of H_2.Explanation / Answer
1) Mg (s) + 2 HCl (aq) --------> MgCl 2 (aq) + H 2 (g)
That is 1 mole of Mg reacts with 2 moles of HCl to give 1 mole of MgCl2 and 1 mole of H2 gas.
2)
Mass of Mg reacts = 0.015 g
Molar mass of Mg = 24.3 g/mol
Number of moles of Mg = (0.015 g)/(24.3 g/mol) = 0.00062 mol.
According to stoichiometry 1 mole of Mg produces 1 mole H2,
0.00062 mol Mg produces 0.00062 mol of H2 (g).
According to ideal gas equation
PV = nRT
Where P = final pressure, n= number of moles of H2
R = 8.314 dm^3 kPa mol^-1K^-1 ;
T = final temparature = 321.9 K{refer answer of 3rd question }
V = nRT /P = (0.00062 mol×8.314×321.9)/108.7 = 0.0152 dm^3
1 dm^3 = 1000 ml
So 0.0152 dm^3 = 15.2 ml
3)
According to gas law at constant volume , as temparature increases pressure also increases.
Thus P/T = constant
So P1/T1 = P2/T2
Where P1 & T1 are initial pressure & temperature .
P2 & T2 are final pressure & temperature.
T2 = (P2×T1)/P1 = (108.7 kPa × 297 K)/100.3 kPa
= 321.9 K = 48.90C
4)
Total initial pressure = pAir
Total final pressure Ptot = pH2 + pAir.
where pH2 is partial pressure of H2 = Ptot - pAir = 108.7 -100.3 =
= 8.4 kPa.
5)
Combined gas law is
P1 V1/T1 = P2 V2 / T2
1 represent STP condition.(P1 = 101.325 kPa)
V1 = (P2 V2 T1)/P1T2 = (108.7×(153- 5)×273)/(101.325×321.9)
134.65 ml
6)
Mass of Mg reacts = 0.015 g
Molar mass of Mg = 24.3 g/mol
Number of moles of Mg = (0.015 g)/(24.3 g/mol) = 0.00062 mol.
According to stoichiometry 1 mole of Mg produces 1 mole H2,
0.00062 mol Mg produces 0.00062 mol of H2 (g).
0.00062 mol H2 = 0.00062 ×1.008 = 0.00062 g H2.
7)
0.00062 mol occupies 15.2 ml
So 1 mole occupies = 15.2/0.00062 = 24,516.12 ml = 24.52 L
So molar volume = 24.52 mol/ L
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