A student following the procedure described in this module collected the follwin
ID: 726648 • Letter: A
Question
A student following the procedure described in this module collectedthe follwing data:
1. Mass Mg, m= 0.0243, initial syringe volume, ml=1.0, final syringe
volume,ml= 26.5, barometric pressure, torr= 754, temperature,k= 298.
Calculate the gas law constant. R=pv/nt Find the R value
2. What would be the volume gas produced by the reaction of .243g of
magnesium metal and collected at 750 torr and 298k? Use the value R in
the previous problem.
3. what would be the effect on the results of mistakenly adding 2.43 x 10^-3g Mg to 4.0mL of 0.1M HCl, rather than to 4.0mL of 1.0M HCl as instructed?
4. If temperature and pressure remain constant, how does the volume of gas sample vary with the number of moles of gas present?
Explanation / Answer
Ans: 1. PV = nRT for R R = (750torr)(26.5L) / [(1mole)(298K)] = 66.96torrL/moleK 2. Solve PV = nRT for R. R = PV / nT Substitute in appropriate values for 1mole of gas at STP. If the problem specified a particular unit for volume use that. Since none was specified, choose L. R = (750torr)(26.5L) / [(1mole)(298K)] = 66.96torrL/moleK You've listed two different values for mass of Mg. It looks like you're intended to change 0.243g to 0.0100moles by inspection since the molar mass for Mg = 24.3. Plug and chug: V = nRT / P = (0.0100mole) x (66.96torrL/moleK) x (298K) / 750torr = 0.266L 3. Interesting question. 2.43e-3gMg [ 1moleMg / 24.3gMg ] x [1mole H2 / 1mole Mg] = 1e-4molesH2 produced. 0.004L x 0.1M HCl = 0.0004molesHCl x [1mole H2 / 2 moles HCl ] = 2e-4moles H2 produced. It turns out the Mg is the limiting reactant. That means that adding more HCl would not increase the amount of H2 gas produced. The acid left over in the reaction container will be more dilute, which could be a safety benefit. 4. PV = nRT V = nRT / P RT and P are constant so V = kn where k is a constant: k=RT/P V varies directly as n.
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