Question

How would you calculate the mass in grams of hydrogen chloride produced when 4.9 L of molecular hydrogen at STP reacts with an excess of chlorine gas.?

Answer 1

`"16 g"`

Explanation:

The important thing to notice here is that the reaction takes place at STP conditions, which are defined as a pressure of `"100 kPa"` and a temperature of `0^@"C"`.

Moreover, at STP one mole of any ideal gas occupies exactly `"22.7 L"` - this is known as the molar volume of a gas at STP.

Since all the gases are at the same conditions for pressure and temperature, the mole ratios become volume ratios.

To prove this, use the ideal gas law equation to write the number of moles of hydrogen gas and of chlorine gas as

`PV = nRT implies n = (PV)/(RT)`

For hydrogen, you would have

`n_"hydrogen" = (P * V_"hydrogen")/(RT)`

and for chlorine you have

`n_"chlorine" = (P * V_"chlorine")/(RT)`

Thus, the mole ratio between hydrogen and chlorine will be

`n_"hydrogen"/n_"chlorine"= (color(red)(cancel(color(black)(P))) V_"hydronge")/color(red)(cancel(color(black)(RT))) * color(red)(cancel(color(black)(RT)))/(color(red)(cancel(color(black)(P))) * V_"chlorine") = V_"hydrogen"/V_"chlorine"`

The same principle applies to the mole ratio that exists between hydrogen and hydrogen chloride.

So, the balanced chemical equation for this reaction is

`"H"_text(2(g]) + "Cl"_text(2(g]) -> color(blue)(2)"HCl"_text((g])`

Notice that you have a `1:color(blue)(2)` mole ratio between hydrogen gas and hydrogen chloride.

This means that the reaction will produce twice as many moles as you the number of moles of hydrogen gas that reacts.

Use the volume ratio to find what volume of hydrogen chloride will be produced by the reaction

`4.9color(red)(cancel(color(black)(" L H"_2))) * (color(blue)(2)" L HCl")/(1color(red)(cancel(color(black)(" L H"_2)))) = "9.8 L HCl"`

Now use the molar volume to find how many moles you'd get in this volume of gas at STP

`9.8color(red)(cancel(color(black)(" L HCl"))) * "1 mole HCl"/(22.7color(red)(cancel(color(black)(" L HCl")))) = "0.4317 moles HCl"`

Finally, use hydrogen chloride's molar mass to find how many grams would contain this many moles

`0.4317color(red)(cancel(color(black)("moles HCl"))) * "36.461 g"/(1color(red)(cancel(color(black)("mole HCl")))) = "15.74 g"`

Rounded to two sig figs, the answer will be

`m_"HCl" = color(green)("16 g")`