Question #79f4c

1 Answer
Nov 6, 2015

Answer:

#1.7%#

Explanation:

A very important thing to recognize here is that the gas collected in the tube will contain both hydrogen gas and water vapor.

Usually, the problem provides you with the values of water's vapor pressure at the temperature at which the experiment takes place. Since this is not the case here, you can look it up yourself

http://www.endmemo.com/chem/vaporpressurewater.php

So, the water vapor pressure at #21.0^@"C"# can be approximated to #"18.6 mmHg"#. Keep this in mind.

Now, magnesium will react with hydrochloric acid according to the balanced chemical equation

#"Mg"_text((s]) + 2"HCl"_text((aq]) -> "MgCl"_text(2(aq]) + "H"_text(2(g]) uarr#

Use magnesium's molar mass to determine how many moles you have in that sample

#0.0375color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305color(red)(cancel(color(black)("g")))) = "0.001543 moles Mg"#

Since the hydrochloric acid is in excess, you can say that the reaction will produce

#0.001543color(red)(cancel(color(black)("moles Mg"))) * ("1 mole H"_2)/(1color(red)(cancel(color(black)("mole Mg")))) ="0.001543moles H"_2#

Now, to get the pressure of the hydrogen gas produced by the reaction, use the pressure of the room and the vapor pressure of the water vapor to determine the pressure of the hydrogen.

Notice that the pressure of the room is given in inHg, which means that you're going to have to comvert it to mmHg

#29.85color(red)(cancel(color(black)("inHg"))) * "25.4 mmHg"/(1color(red)(cancel(color(black)("inHg")))) = "758.2 mmHg"#

The pressure of the hydrogen will thus be

#P_"hydrogen" = "758.2 mmHg" - "18.6 mmHg" = "739.6 mmHg"#

The ideal gas law equation looks like this

#color(blue)(PV = nRT)" "#, where

#P# - the pressure of the gas
#V# - the volume it occupies
#n# - the number of moles of gas
#R# - the universal gas constant
#T# - the temperature expressed in Kelvin

Rearrange this equation to solve for #R#, then plug in your values - do not forget to convert the temperature, volume, and pressure to Kelvin, liters, and atm, respectively!

#R = (PV)/(nT)#

#R = (739.6/760"atm" * 38.93 * 10^(-3)"L")/("0.001543 moles" * (273.15 + 21.0)"K") = 0.0835("atm" * "L")/("mol" * "K")#

The accepted value of #R# is #0.0821("atm" * "L")/("mol" * "K")#, which means that the percent error will be

#color(blue)("% error" = ( |"experimental value" - "actual value"|)/"actual value" xx 100)#

#"% error" = ( |0.0821 - 0.0835|)/0.0821 xx 100 = color(green)(1.7%)#