Name Gases Lab Quiz Mg(s)+2HCI(aq) MgCla(aq) 2(8) A student reacted with 0.094 g
ID: 1041393 • Letter: N
Question
Name Gases Lab Quiz Mg(s)+2HCI(aq) MgCla(aq) 2(8) A student reacted with 0.094 g of Magnesium with excess 3.0 M hydrochloric acid, an reaction stopped, the voltme of gas collected was 0.0485 Lat 0.986 atm and 296 K. a. Calculate the actual number of moles of hydrogen (H2) produced. Assume the vapor pressure of b. d collected the displacement of water in a gas collection tube. When the water at 296 K is 21.1 torr. Calculate the number of moles of hydrogen (H2) that would have been produced if all of the magnesium had reacted Calculate the percent yield of hydrogen in the reaction. The magnesium chloride formed in the above reaction remained in solution. Draw particle level diagrams to show how this may appear when magnified. c. d.Explanation / Answer
1)To calculate actual number of moles of hydrogen produced:
Using ideal gas law PV = nRT
where P= pressure, V=volume, n=number of moles of a gas and T= temperature. R=the molar gas constant, which is 8.314472. The gas constant lets you work with the standard units of Kelvins for temperature, moles of gas amount, pressure in pascals and volume in cubic meters.
Multiply the pressure by 101,325 to convert the pressure to the International System of Units pascal (Pa)
PV=nRT
n = PV / RT
n = 99906.45*4.8*10^-5 /8.31441*296
n= 1.948*10^-3 moles of hydrogen
2) NO.of moles of hydrogen = total no. Of moles of Mg*1M H2/1mole of Mg
= 9.094 *1M/24.30
= 3.86*10^-3 moles of hydrogen produced if all the Mg had been reacted.
3) %yield of Hydrogen in the reaction= 1.948*10^-3/3.86*10^-3*100
= 50.35%
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