1.9 mol HCl and 3.8 mol NaOH react according to the equation
HCl + NaOH −→ NaCl + H2O .
Calculate the amount in moles of NaCl
formed.
Answer in units of mol
15.4
To determine the amount in moles of NaCl formed, we need to use the stoichiometry of the balanced equation.
From the balanced equation:
1 mol HCl reacts with 1 mol NaOH to form 1 mol NaCl.
Given:
1.9 mol HCl
3.8 mol NaOH
Since the ratio of HCl to NaOH is 1:1, we can use either the amount of HCl or NaOH given. In this case, we will use the amount of NaOH.
Therefore, 3.8 mol NaOH will react to form 3.8 mol NaCl.
To calculate the amount in moles of NaCl formed, we need to determine the limiting reactant in the reaction. The limiting reactant is the one that is completely consumed and therefore determines the maximum amount of product that can be formed.
To find the limiting reactant, we compare the stoichiometric ratios of the reactants (HCl and NaOH) to the product (NaCl) based on the balanced equation.
From the balanced equation:
1 mol HCl reacts with 1 mol NaCl
Now, let's calculate the moles of NaCl that can be formed by each reactant:
For HCl:
Moles of NaCl = moles of HCl = 1.9 mol
For NaOH:
Moles of NaCl = moles of NaOH = 3.8 mol
Since the stoichiometric ratio of both reactants to NaCl is 1:1, we can conclude that the limiting reactant is HCl (with 1.9 mol). This means that 1.9 mol of NaCl will be formed in this reaction.
Therefore, the amount in moles of NaCl formed is 1.9 mol.
This is limiting reagent problem.
1.9 mol HCl can form 1.9 mol NaCl if it has all of the NaOH it needs (It will need 1.8 mol NaOH).
3.8 mol NaOH can form 3.8 mol NaCl if it has all of the HCl needed(it would need, of course, 3.8 mol HCl. It doesn't have that much).
Therefore, the 1.9 mol HCl will form 1.9 mol NaCl and it will use 1.9 mol NaOH in the process. There will be some NaOH remaining unreacted. 3.8-1.9 = 1.9 moll NaOH unreacted.