if 50 grams of sodium hydroxide mixed with 25 grams of hydrochloric acid how many grams of salt can be formed?
You have a limiting reagent (LR) problem here because amounts are given for BOTH reactants.
NaOH + HCl ==> NaCl + H2O
mols NaOH = grams/molar mass= ?
mols HCl ==> grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols NaOH to mols NaCl.
Do the same and convert mols HCl to mols NaCl.
It is likely the two values will not agree which means one of them is wrong; the correct value in LR problems is ALWAYS the smaller one and the reagent responsible for that is the LR.
using the smaller value convert to g NaCl. g NaCl = mols NaCl x molar mass NaCl.
To determine how many grams of salt can be formed when 50 grams of sodium hydroxide is mixed with 25 grams of hydrochloric acid, we need to understand the chemical reaction that occurs between these two substances.
Sodium hydroxide (NaOH) and hydrochloric acid (HCl) react to form sodium chloride (NaCl) and water (H2O) in a chemical reaction known as a neutralization reaction.
The balanced chemical equation for this reaction is:
NaOH + HCl -> NaCl + H2O
The equation tells us that one mole of sodium hydroxide reacts with one mole of hydrochloric acid to produce one mole of sodium chloride and one mole of water.
To calculate the grams of salt formed, we need to determine the limiting reactant. The limiting reactant is the reactant that is used up first and limits the amount of product that can be formed in a chemical reaction.
To do this, we need to convert the masses of sodium hydroxide and hydrochloric acid to moles by using their respective molar masses. The molar mass of sodium hydroxide (NaOH) is 39.997 grams per mole, and the molar mass of hydrochloric acid (HCl) is 36.461 grams per mole.
The number of moles of sodium hydroxide can be calculated as follows:
moles of NaOH = mass of NaOH / molar mass of NaOH
moles of NaOH = 50 grams / 39.997 grams per mole
Similarly, the number of moles of hydrochloric acid can be calculated as:
moles of HCl = mass of HCl / molar mass of HCl
moles of HCl = 25 grams / 36.461 grams per mole
Now, if we compare the moles of sodium hydroxide and hydrochloric acid, we can see which is the limiting reactant. The reactant with the smaller number of moles is the limiting reactant.
Assuming we have excess hydrochloric acid:
moles of NaOH : moles of HCl = 50 grams / 39.997 grams per mole : 25 grams / 36.461 grams per mole
Calculating these ratios will give us the molar ratios of the reactants, allowing us to determine the limiting reactant.
Now, once we know the limiting reactant, we can calculate the moles of salt (sodium chloride) formed using the stoichiometry of the balanced chemical equation.
From the balanced equation, we know that one mole of sodium hydroxide reacts to form one mole of sodium chloride. Therefore, the moles of salt formed will be equal to the moles of sodium hydroxide used (if it is the limiting reactant).
Finally, to determine the grams of salt formed, we use the molar mass of sodium chloride, which is 58.443 grams per mole.
grams of salt = moles of salt × molar mass of salt
Let's plug in the numbers to get the final answer.