Please help me with these questions:
1. Consider the equation:2Na+ 2H2O-->2NaOH +H2
-If 92.0g of sodium is reacted with 76.0g of water until the reaction goes to completion, which reactant will remain and in what quantity?
A: I know that H20 is the reactant that is in excess with 4.2 moles whereas Na has 4.0 moles.-But how do you find the quantity?
2. 12.5 ml of 0.280M HNO3 and 5.0mL of 0.920M KOH are mixed. To make the resulting solution neutral, which of the following should be added?
-1.1 mol HCl
-1.69 mL of 0.650M HBr
-0.55 mmol Mg(OH)2
-2.2mL of 0.50M LiOH
-none of the above; already neutral
You work it the same way you do any stoichiometry problem.
You know Na is the limiting reagent, it will be completely consumed; therefore, how much H2O must be used according to the equation? It must be
4.00moles Na x (2 moles H2O/2 moles Na) = 4.00 x (1) = 4.00 moles H2O. You had 4.2; therefore, 0.2 mols H2O must remain un-reacted. How many grams is that?
0.2 x 18 = 3.6 grams.
76.0-3.60 = ?? that didn't react.(amount not consumed).
Now that you see how to do that you probably can do the second problem. moles HNO3 - moles KOH = moles in excess.
1. To find the quantity of the reactant that remains, you need to determine which reactant is the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed and determines the amount of product formed.
First, calculate the moles of each reactant:
moles of Na = 92.0g / molar mass of Na
moles of H2O = 76.0g / molar mass of H2O
Next, you need to determine the stoichiometric ratio between Na and H2O in the balanced equation. From the equation, you can see that 2 moles of Na react with 2 moles of H2O to produce 2 moles of NaOH and 1 mole of H2.
Now, compare the moles of Na and H2O to see which reactant is limiting:
moles of Na / stoichiometric ratio of Na = moles of NaOH (products) produced
moles of H2O / stoichiometric ratio of H2O = moles of NaOH (products) produced
The reactant that produces fewer moles of NaOH is the limiting reactant. The reactant that produces more moles of NaOH is in excess.
In this case, if you find that the moles of Na are fewer than the moles of H2O, then Na is the limiting reactant. The quantity of Na remaining can be found by subtracting the moles of Na that reacted from the initial moles of Na used.
2. To determine which substance should be added to make the resulting solution neutral, you need to perform a stoichiometric calculation.
First, calculate the number of moles of HNO3 and KOH used:
moles of HNO3 = volume (in L) x molarity = 12.5 mL x (0.280 mol/L)
moles of KOH = volume (in L) x molarity = 5.0 mL x (0.920 mol/L)
Next, determine the stoichiometric ratio between HNO3 and KOH in the balanced equation. From the equation, you can see that 1 mole of HNO3 reacts with 1 mole of KOH to produce 1 mole of water.
Now, compare the moles of HNO3 and KOH to see which one is in excess. The substance that is in excess will determine what needs to be added to make the solution neutral.
In this case, if you find that the moles of HNO3 are greater than the moles of KOH, then KOH is the limiting reactant. Therefore, to neutralize the solution, you need to add an acid.
Looking at the given options, you should add 1.1 mol HCl, as it is an acid that can react with the excess KOH to form water and a salt, resulting in a neutral solution.
1. To find the quantity of the remaining reactant, you need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed in the reaction and determines the maximum amount of product that can be formed.
To determine the limiting reactant, you can convert the given masses of sodium and water to moles using their respective molar masses. The molar mass of sodium (Na) is 22.99 g/mol and the molar mass of water (H2O) is 18.02 g/mol.
For sodium (Na):
- Mass = 92.0 g
- Moles = Mass/Molar mass = 92.0 g/22.99 g/mol ≈ 4.00 moles
For water (H2O):
- Mass = 76.0 g
- Moles = Mass/Molar mass = 76.0 g/18.02 g/mol ≈ 4.22 moles
Since the stoichiometric ratio between sodium (Na) and water (H2O) is 2:2, we can see that both reactants have a stoichiometric ratio of 2 moles. Therefore, the limiting reactant is the one with the lower number of moles, which is sodium (Na) in this case.
To find the quantity of the remaining reactant, we need to calculate the amount of sodium that reacts based on the balanced equation. According to the equation, 2 moles of sodium react with 2 moles of water to produce 2 moles of sodium hydroxide (NaOH) and 1 mole of hydrogen gas (H2).
Since we know that 4.00 moles of sodium are present, and the stoichiometric ratio is 2 moles of sodium to 2 moles of water, it means that 4.00 moles of sodium will react with 4.00 moles of water. This indicates that all of the water will be consumed in the reaction, and no water will remain.
Therefore, the reactant that remains is sodium (Na), and its quantity will be the excess amount that was not consumed. In this case, the remaining quantity of sodium will be 4.00 moles.
2. To determine which substance should be added to make the resulting solution neutral, we need to consider the reaction that occurs when an acid and a base are mixed. The acid donates a proton (H+), and the base accepts the proton.
In this case, we're mixing HNO3 (nitric acid) and KOH (potassium hydroxide). The reaction between nitric acid and potassium hydroxide forms water (H2O) and a salt (KNO3).
The balanced equation for the reaction is as follows:
HNO3 + KOH → H2O + KNO3
To make the resulting solution neutral, we want the acid and base to react completely, resulting in an equal number of moles of H+ and OH- ions, which combine to form water.
Comparing the moles of acid and base, we can calculate the moles of HNO3 and KOH using their concentrations and volumes:
- Moles of HNO3 = concentration (mol/L) × volume (L) = 0.280 mol/L × 0.0125 L
- Moles of KOH = concentration (mol/L) × volume (L) = 0.920 mol/L × 0.0050 L
By comparing the moles of acid and base, we can determine that the moles of KOH are greater than the moles of HNO3. Thus, after the reaction, there will be leftover KOH.
To find which substance should be added to make the solution neutral, we should search for a compound that is an acid and can react with KOH to form water. Among the options provided, HCl (hydrochloric acid) is the suitable choice, as it can react with KOH to produce water (H2O) and form KCl (potassium chloride).
Therefore, the correct option is 1.1 mol of HCl, which should be added to make the solution neutral.