For each of the following acid-base reactions, calculate how many grams of each acid are necessary to completely react with and neutralize 2.1g of the base.
1) HCl(aq)+NaOH(aq)-->H2O(l)+NaCl(aq)
2) 2HNO3(aq)+Ca(OH)2(aq)-->2H20(l)+Ca(NO3)2(aq)
3) H2SO4(aq)+2KOH(aq)-->2H2O(l)+K2SO4(aq)
These are regular stoichiometry problems. Here is an example of a stoichiometry problem I've posted. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the number of grams of acid needed to neutralize a given amount of base, we need to use the stoichiometry of the balanced chemical equation.
1) HCl(aq) + NaOH(aq) --> H2O(l) + NaCl(aq)
From the equation, we see that the molar ratio between HCl and NaOH is 1:1. This means that 1 mole of HCl reacts with 1 mole of NaOH.
To calculate the number of moles of HCl needed, we can use the molar mass of HCl:
Molar mass of HCl = 1.00784 g/mol (H) + 35.453 g/mol (Cl) = 36.46084 g/mol
Number of moles of HCl needed = (2.1 g base) / molar mass of HCl
Now we can calculate the amount of HCl in grams:
Mass of HCl = Number of moles of HCl needed * molar mass of HCl
2) 2HNO3(aq) + Ca(OH)2(aq) --> 2H2O(l) + Ca(NO3)2(aq)
From the equation, we see that the molar ratio between HNO3 and Ca(OH)2 is 2:1. This means that 2 moles of HNO3 react with 1 mole of Ca(OH)2.
To calculate the number of moles of HNO3 needed, we can use the molar mass of HNO3:
Molar mass of HNO3 = 1.00784 g/mol (H) + 14.0067 g/mol (N) + 3(15.999 g/mol) = 63.012 g/mol
Number of moles of HNO3 needed = (2.1 g base) / molar mass of HNO3
Now we can calculate the amount of HNO3 in grams:
Mass of HNO3 = Number of moles of HNO3 needed * molar mass of HNO3
3) H2SO4(aq) + 2KOH(aq) --> 2H2O(l) + K2SO4(aq)
From the equation, we see that the molar ratio between H2SO4 and KOH is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of KOH.
To calculate the number of moles of H2SO4 needed, we can use the molar mass of H2SO4:
Molar mass of H2SO4 = 2(1.00784 g/mol (H)) + 32.065 g/mol (S) + 4(15.999 g/mol) = 98.087 g/mol
Number of moles of H2SO4 needed = (2.1 g base) / molar mass of H2SO4
Now we can calculate the amount of H2SO4 in grams:
Mass of H2SO4 = Number of moles of H2SO4 needed * molar mass of H2SO4
To calculate the grams of each acid necessary to completely react with and neutralize 2.1g of the base, we need to determine the molar ratio between the acids and the base in each reaction.
1) HCl(aq) + NaOH(aq) --> H2O(l) + NaCl(aq)
The balanced equation shows that one mole of HCl reacts with one mole of NaOH. We need to find the molar masses of HCl and NaOH to calculate the grams.
- Molar mass of HCl = 1.008 g/mol (for hydrogen) + 35.453 g/mol (for chlorine) = 36.461 g/mol.
- Molar mass of NaOH = 22.990 g/mol (for sodium) + 15.999 g/mol (for oxygen) + 1.008 g/mol (for hydrogen) = 39.997 g/mol.
The molar ratio is 1:1, so to calculate the grams of HCl required:
- grams of HCl = molar mass of HCl * moles of HCl.
We can determine the moles of HCl using the molar mass of NaOH and the given mass of the base.
- Moles of NaOH = given mass of NaOH / molar mass of NaOH.
- Moles of HCl = Moles of NaOH (since the molar ratio is 1:1).
Then, we can calculate the grams of HCl:
- grams of HCl = molar mass of HCl * Moles of HCl.
2) 2HNO3(aq) + Ca(OH)2(aq) --> 2H20(l) + Ca(NO3)2(aq)
This balanced equation shows that two moles of HNO3 react with one mole of Ca(OH)2. We need to find the molar masses of HNO3 and Ca(OH)2 to calculate the grams.
- Molar mass of HNO3 = 1.008 g/mol (for hydrogen) + 14.007 g/mol (for nitrogen) + 3 * 15.999 g/mol (for oxygen) = 63.014 g/mol.
- Molar mass of Ca(OH)2 = 40.078 g/mol (for calcium) + 2 * 15.999 g/mol (for oxygen) + 2 * 1.008 g/mol (for hydrogen) = 74.092 g/mol.
The molar ratio is 2:1, so to calculate the grams of HNO3 required:
- grams of HNO3 = molar mass of HNO3 * moles of HNO3.
We can determine the moles of HNO3 using the molar mass of Ca(OH)2 and the given mass of the base.
- Moles of Ca(OH)2 = given mass of Ca(OH)2 / molar mass of Ca(OH)2.
- Moles of HNO3 = 2 * Moles of Ca(OH)2 (since the molar ratio is 2:1).
Then, we can calculate the grams of HNO3:
- grams of HNO3 = molar mass of HNO3 * Moles of HNO3.
3) H2SO4(aq) + 2KOH(aq) --> 2H2O(l) + K2SO4(aq)
This balanced equation shows that one mole of H2SO4 reacts with two moles of KOH. We need to find the molar masses of H2SO4 and KOH to calculate the grams.
- Molar mass of H2SO4 = 2 * 1.008 g/mol (for hydrogen) + 32.06 g/mol (for sulfur) + 4 * 15.999 g/mol (for oxygen) = 98.089 g/mol.
- Molar mass of KOH = 39.098 g/mol (for potassium) + 15.999 g/mol (for oxygen) + 1.008 g/mol (for hydrogen) = 56.105 g/mol.
The molar ratio is 1:2, so to calculate the grams of H2SO4 required:
- grams of H2SO4 = molar mass of H2SO4 * moles of H2SO4.
We can determine the moles of H2SO4 using the molar mass of KOH and the given mass of the base.
- Moles of KOH = given mass of KOH / molar mass of KOH.
- Moles of H2SO4 = (1/2) * Moles of KOH (since the molar ratio is 1:2).
Then, we can calculate the grams of H2SO4:
- grams of H2SO4 = molar mass of H2SO4 * Moles of H2SO4.
By following these steps, you can calculate the grams of each acid required to completely neutralize the base in each reaction.