H2(g) + Cl2(g) HCl(g)

Consider the unbalanced equation above. 3.95 L of H2 and 1.00 L of Cl2 are reacted under the same conditions of temperature and pressure.
(a) How many liters of HCl will be produced?
(b) How many liters of reactant are in excess?

i found part a to be 2.00 L what do i need to do to find part b

To find the answer to part (b), you first need to balance the equation. The unbalanced equation is:

H2(g) + Cl2(g) -> HCl(g)

To balance the equation, you need to ensure that the number and type of atoms are the same on both sides.

By inspection, you can see that the equation is already balanced as it is written. There is one hydrogen atom on the left side and one hydrogen atom on the right side. Similarly, there are two chlorine atoms on both sides.

Now, you need to determine the stoichiometry of the reaction. From the balanced equation, you can see that the ratio of H2 to HCl is 1:2. This means that for every 1 mole of H2 consumed, 2 moles of HCl are produced.

To find the number of moles of H2 consumed, you can use the ideal gas law:

PV = nRT

Assuming constant temperature and pressure, the number of moles can be calculated using the ideal gas law formula:

n = PV / RT

Given that the volume of H2 is 3.95 L and the volume of Cl2 is 1.00 L, you can substitute these values into the equation along with the appropriate gas constant (R) and temperature (T). Let's assume the temperature is 298 K and R is 0.0821 L.atm/mol.K:

n(H2) = (3.95 L) / (0.0821 L.atm/mol.K * 298 K)
= 0.1626 moles

According to the balanced equation, for every 1 mole of H2 consumed, 2 moles of HCl are produced. Therefore, the number of moles of HCl produced can be calculated as:

n(HCl) = 2 * n(H2)
= 2 * 0.1626 moles
= 0.3252 moles

To convert moles to liters, you can use the ideal gas law again, rearranged to solve for volume:

V = nRT / P

Assuming the same pressure (P) as before, you can substitute the values into the formula:

V(HCl) = (0.3252 moles) * (0.0821 L.atm/mol.K * 298 K) / (P)
= 8.07 L / (P)

Since the reaction occurs under the same conditions of temperature and pressure, you can assume that the pressure is constant. Therefore, the 0.3252 moles of HCl will occupy 8.07 L, and the answer to part (a) is 8.07 L.

To find the answer to part (b), you need to compare the volumes of the reactants (H2 and Cl2) used. In this case, the volume of H2 (3.95 L) is bigger than the volume of Cl2 (1.00 L), which means that H2 is in excess.

Therefore, the answer to part (b) is that H2 is the reactant in excess.

2.000L for part a is correct.

H2 + Cl2 ==> 2HCl

When working with gases remember that you can take a short cut and use L directly as mols IF everything stays at the same T and P. So I do this.
...........H2 + Cl2 ==> 2HCl
I.........3.95..1.00....0
C........-1.00.-1.00...1.00
E.........2.95...0.....1.00

So you have used 1.00 mol of the 3.95 mols to leave 2.95 mols H2 remaining un-reacted. Of course that's the same as
3.95 L H2 initially less 1.00 L reacted = 2.95 L H2 remaining un-reacted.