Gas Law problems with stoichiometry

Consider the following reaction : $\ce{H2 + Cl2 -> 2HCl}$

A) If you have 15.00 liters of hydrogen reacting with 10.00 liters of chlorine at STP, how many liters of HCl would you produce?

B) What would be the limiting reagent?

C) How many moles of excess reagent would remain?

Molarity = moles/Liters to check for $\ce{H2}$: M = 2/15.00 = .13334 to check for $\ce{Cl2}$: M = (35.45 x 2)/10.00 = 7.09

But I do not think that is correct. can somebody show me the correct way to do it.

• Please see my comment on your other question. – jonsca Dec 3 '13 at 7:13

Avogadro's law says that $V/n = \text{constant}$ at constant temperature and pressure. That means your volumes can be used "as if" they were moles in the chemical reaction stoichiometry. So, if the mole ratio is 1:1 for H$_2$ and Cl$_2$ reacting, then the limiting reagent is obviously the smaller volume of 10.0 L for Cl$_2$.

For (b) you can easily calculate the volume of HCl since 1 mole of Cl$_2$ produces 2 moles of HCl. So, you get 20.0 L of HCl.

For (c) the excess reagent is the H$_2$. Since it reacts with 10.0 L of Cl$_2$ in a 1:1 mole ratio, then you have 15.0L - 10.0 L = 5.0 L of H$_2$ left over. Since this all ends up at STP, you can use the ideal gas law to calculate the moles of H$_2$ left over:

$n_{H_2} = \displaystyle \frac{p V}{R T} = \frac{(1 \text{atm})(5.0 \text{L})}{R \times 273.15 \text{K}} = 0.2230749373505515$ mol

Don't forget sig. figs., which in this case is determined by the 2 sig. figs. in the volume of 5.0 L, giving you

$n_{H_2} = 0.22$ mol

For Review check out: gas laws: http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases

If you have a mac, check out PhySyCalc. It will automatically do all the unit conversions in your calculations.

https://itunes.apple.com/us/app/physycalc/id644105995

You do not need the molar masses to do this question. The ideal gas law includes a relationship between volume $V$ and moles $n$ assuming pressure and temperature are constant (they are at STP).

$$PV=nRT$$ $$n=\dfrac{PV}{RT}$$

At STP, 1 mol of any gas occupies 22.4L. Using this knowledge, you can then convert each volume of gas to moles and then determine the limiting reactant using the 2:1 molar ratio of product to reactant.

As for part a, the amount of HCl produced would be dependent on the limiting reactant.

Part c could then be answered by subtracting the number of moles of limiting reactant from the excess reactant.

You only need to know that at STP $V_m=24.4\:\mathrm{L\:mol^{-1}}$, and that $n$ (number of moles of of a compound or molecule) is equal to $\frac{V_{gas}}{V_m}$.

Find limiting and excess reagents using stoichiometric ratios (in this case it is 1:1), therefore $\ce{H2}$ is limiting the limiting reagent. Use stoichiometry to find amount of $\ce{Cl2}$ reacted, and then subtract the initial number of moles from the final number of moles.