Identify the excess reactant (HCl or NaOH) and its molar concentration when 1.55 g of NaOH is stirred into 150.0 mL of 0.250 M HCl.
mols NaOH = grams/molar mass = 0.03875
mols HCl = M x L = 0.0375
The difference will tell you that NaOH is in excess and by how many mols.
Then M NaOH = mols/L.
To identify the excess reactant, we need to compare the amount of each reactant present in the given solution to the stoichiometric ratio of the balanced equation.
First, let's calculate the number of moles of NaOH and HCl using their respective molar masses:
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Number of moles of NaOH = mass / molar mass = 1.55 g / 40.00 g/mol ≈ 0.0388 mol
Molar mass of HCl = 1.01 g/mol + 35.45 g/mol = 36.46 g/mol
Number of moles of HCl = concentration × volume = 0.250 M × 0.1500 L = 0.0375 mol
Now, let's determine the stoichiometric ratio of the balanced equation between NaOH and HCl:
NaOH + HCl -> NaCl + H2O
1 mol 1 mol
Based on the balanced equation, the ratio between NaOH and HCl is 1:1.
Since the ratio is 1:1, we can see that the limiting reactant will be whichever reactant has a smaller number of moles. In this case, HCl has 0.0375 moles, whereas NaOH has 0.0388 moles.
Hence, HCl is the limiting reactant with a molar concentration of 0.250 M. NaOH is the excess reactant.
To identify the excess reactant, we need to determine which reactant will be left over after the reaction is complete.
Let's start by writing the balanced equation for the reaction between HCl and NaOH:
HCl + NaOH -> NaCl + H2O
From the balanced equation, we can see that the molar ratio between HCl and NaOH is 1:1. This means that for every 1 mole of HCl, 1 mole of NaOH is required for complete reaction.
First, let's calculate the number of moles of NaOH used in the reaction:
Mass of NaOH = 1.55 g
Molar mass of NaOH = 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 39.00 g/mol
Number of moles of NaOH = Mass of NaOH / Molar mass of NaOH
Number of moles of NaOH = 1.55 g / 39.00 g/mol ≈ 0.0397 mol
Next, let's calculate the number of moles of HCl required for complete reaction with the given volume of NaOH solution:
Volume of HCl = 150.0 mL = 150.0 cm³
Molar concentration of HCl = 0.250 M
Number of moles of HCl = Molar concentration of HCl × Volume of HCl (converted to liters)
Number of moles of HCl = 0.250 mol/L × 150.0 cm³ × (1 L / 1000 cm³)
Number of moles of HCl = 0.0375 mol
Comparing the moles of NaOH (0.0397 mol) to the moles of HCl (0.0375 mol), we can see that the HCl is the limiting reactant because there is a smaller amount of moles available.
Therefore, HCl is the limiting reactant, and NaOH is the excess reactant.
To determine its molar concentration, we need to calculate the number of moles of excess NaOH remaining after the reaction. Since HCl is the limiting reactant, all of the HCl will be consumed, and some NaOH will be left.
Number of moles of excess NaOH = Moles of NaOH used in the reaction - Moles of NaOH required for complete reaction
Number of moles of excess NaOH = 0.0397 mol - 0.0375 mol ≈ 0.0022 mol
Finally, let's calculate the molar concentration of the excess NaOH:
Volume of excess NaOH = Volume of NaOH used in the reaction
Volume of NaOH = 150.0 mL = 150.0 cm³
Molar concentration of excess NaOH = Number of moles of excess NaOH / Volume of excess NaOH (converted to liters)
Molar concentration of excess NaOH = 0.0022 mol / (150.0 cm³ × (1 L / 1000 cm³))
Molar concentration of excess NaOH ≈ 0.0147 M
So, the excess reactant is NaOH, and its molar concentration is approximately 0.0147 M.