Ammonia and gaseious hydrogen chloride combine to form ammonium chloride.
NH3(g) + HCl(g) = NH4Cl(s) (balanced)
If 4.21 L of NH3(g) at 27 C and 1.02 atm is combined with 5.35 L of HCl at 26 C and 0.998 atm, what mass of NH4Cl will be produced? Which gas is the limiting reactant? Which gas is present in excess?
PLEASE HELP!
Use PV = nRT (once for each gas) to determine moles. Since the reaction is a 1:1 reaction (1 mole NH3 to 1 mole HCl), the reactant with the smaller number of moles will be the limiting reagent and convert that number of moles to grams NH4Cl.
So HCL would be the limiting reactant, right? And NH3 is present in excess. So do I convert 0.218 mol HCl to grams of NH4Cl?
Thanks btw for helping me. :)
no
wery no.
To find the mass of NH4Cl produced and determine which gas is the limiting reactant and which is in excess, you need to follow these steps:
Step 1: Convert the given volumes to moles.
Use the ideal gas law equation, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is temperature in Kelvin.
For NH3:
P1 = 1.02 atm (given)
V1 = 4.21 L (given)
T1 = 27°C = 27 + 273 = 300 K
R = 0.0821 L·atm/(mol·K)
Using PV = nRT, rearrange the equation to solve for n1 (moles of NH3):
n1 = (P1 * V1) / (R * T1)
For HCl:
P2 = 0.998 atm (given)
V2 = 5.35 L (given)
T2 = 26°C = 26 + 273 = 299 K
R = 0.0821 L·atm/(mol·K)
Using PV = nRT, rearrange the equation to solve for n2 (moles of HCl):
n2 = (P2 * V2) / (R * T2)
Step 2: Calculate the stoichiometry.
From the balanced equation, you can see that the stoichiometric mole ratio between NH3 and NH4Cl is 1:1. This means that one mole of NH3 reacts completely to produce one mole of NH4Cl.
Step 3: Determine the limiting reactant.
The limiting reactant is the reactant that is completely consumed first, thereby limiting the amount of product that can be formed. To find the limiting reactant, compare the moles of NH3 (n1) and HCl (n2) calculated in step 1. The reactant that produces the smaller number of moles is the limiting reactant.
Step 4: Calculate the moles of NH4Cl produced.
Since the stoichiometry is 1:1 between NH3 and NH4Cl, the number of moles of NH4Cl produced is equal to the number of moles of the limiting reactant.
Step 5: Convert moles of NH4Cl to mass in grams.
Use the molar mass of NH4Cl to convert moles to grams.
Once you have followed these steps, you will find the mass of NH4Cl produced and identify the limiting reactant and the reactant in excess.