Hi, Mg(s) + 2 HCI(aq) MgCl2(aq) + H2(g) Pressure= ? mass 0.0252g Molar mass 24.3

Question

Hi,

Mg(s) + 2 HCI(aq) MgCl2(aq) + H2(g) Pressure= ? mass 0.0252g Molar mass 24.305g/mol n= 1.03682*10^-3 mol Volume 0.02655L Temp = 296.95K n= 1.03682*10^-3 mol PV = nRT 1) Calculate the pressure of hydrogen in the gas burette (PH) gase gares Result from the lab experiment: -96.89 kPa-2.954 kPa gases = 96.89kPa -93.94 kPa H, O = 2.954kPa 2)Using the experimental data and the ideal gas equation, calculate an experimental value for the gas constant in L kPa mo K n.T = 8.10 L kPa mol. 1K'l 3) If some magnesium remained after the reaction finished, how would this affect your measu ured value of 'R? Why? 4) If you did not consider the partial pressure of the water in the gas burette (PH2o), how would this affect the measured value of 'R'? Why?

Explanation / Answer

Answer –

Q3 ) The measured value of R will be increased, because when there is some Mg remaining in the after complete reaction then there is formation of H2 gas less than accepted, because when Mg metal not completely react then there is less number of moles of Mg reacted and that affect the formation of moles of H2 gas.

When we will get less number of moles of H2 gas then there is the value in the denominator bellow formula is gets low

R = PV/nT

That affect the coming value of R is high that accepted, since denominator is gets less than actual.

Q 4) The measured value of R will be decreased, because when we didn’t consider the partial pressure of water then there is pressure of gas is more than actual, means there is total pressure consider in the calculating value of R in the bellow formula

R = PV/nT

So as the P value increased than actual, so the result value of R will be decrease, since the numerator value increased and hence the result of the ratio gets decreased.

The remaining calculation are correct.

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