can someone explain the stoichiometry going on here...
so the questions is: A solution containing 0.20g of calcium salt is passed through a column containing the hydrogen form of a strong acid cation exchange resin. The resulting solution is neutralized with 35 mL of .1 M NaOH. What is the # of moles of Calcium2+ per gram of calcium salt.
So here was my thinking, whilst trying to set up an equation.
CaCl2 + HCl --> Ca2+ + 2Cl
Ca2+ + 2Cl +NaOH --> Ca(OH)2 +2NaCl
M=mol/L
mol=M/L
mol=(35mL)(0.1 mol/L) / (1000mL) (0.20)
but the answer has is 35*0.1/1000*0.2*2
where does the 2 come from! lol
Ca^2+ + 2Hresin ==> 2H^+ + Ca(resin)2
Then H^+ + NaOH ==> H2O + Na^+
mols H^+ = M x L = (35/1000) x 0.1
mols Ca^2+ = 1/2 mols H^+ or
(35/1000)x 1/2
mols Ca^2+ per gram =
(35/1000) x 1/2 x 1/0.2 = ?
The "2" in the answer comes from the balanced chemical equation for the neutralization reaction between calcium ions (Ca2+) and hydroxide ions (OH-):
Ca2+ + 2OH- --> Ca(OH)2.
In this equation, you can see that 1 mole of Ca2+ reacts with 2 moles of OH-.
To calculate the moles of calcium ions per gram of calcium salt, you need to determine the number of moles of calcium salt first. You can do this using the given mass of calcium salt (0.20 g) and its molar mass. Let's assume the calcium salt is calcium chloride (CaCl2) for this calculation.
The molar mass of CaCl2 is:
Ca: 40.08 g/mol
Cl: 35.45 g/mol x 2 = 70.90 g/mol
So, the molar mass of CaCl2 is 40.08 g/mol + 70.90 g/mol = 111.98 g/mol.
Now, let's calculate the moles of calcium salt (CaCl2):
moles = mass (g) / molar mass (g/mol)
moles = 0.20 g / 111.98 g/mol
After calculating this, you should find that the moles of calcium salt is approximately 0.0018 mol.
Next, we can calculate the moles of calcium ions (Ca2+) from the given equation:
moles of Ca2+ = moles of calcium salt x 2
Using the previously calculated moles of calcium salt:
moles of Ca2+ = 0.0018 mol x 2 = 0.0036 mol
Finally, to find the moles of calcium ions per gram of calcium salt:
moles of Ca2+ per gram of calcium salt = moles of Ca2+ / mass of calcium salt (g)
Plugging in the values:
moles of Ca2+ per gram of calcium salt = 0.0036 mol / 0.20 g
Simplifying this, you get:
moles of Ca2+ per gram of calcium salt ≈ 0.018 mol/g.
So, the correct answer is approximately 0.018 mol/g, assuming the calcium salt is calcium chloride (CaCl2).
To explain the stoichiometry in this problem, let's break down the steps involved:
Step 1: Passage through cation exchange resin:
The calcium salt, which we can assume to be CaCl2, is passed through a column containing a hydrogen form of a strong acid cation exchange resin. In this process, the cations (positively charged ions) in the calcium salt are exchanged with hydrogen ions from the resin. So, calcium cations (Ca2+) are exchanged for hydrogen ions (H+).
CaCl2 (aq) + 2 H+ (resin) -> Ca2+ (aq) + 2 Cl- (aq)
Step 2: Neutralization with NaOH:
The resulting solution, containing Ca2+ ions, is then neutralized with 35 mL of 0.1 M NaOH. The reaction between Ca2+ and NaOH produces calcium hydroxide (Ca(OH)2) and sodium chloride (NaCl) through a double displacement reaction.
Ca2+ (aq) + 2 Cl- (aq) + 2 NaOH (aq) -> Ca(OH)2 (s) + 2 NaCl (aq)
Now let's move on to calculating the number of moles of Ca2+ per gram of calcium salt.
The given equation they provided for the calculation is:
mol = (35 mL) (0.1 mol/L) / (1000 mL) (0.20)
However, this equation appears to be missing a factor of 2. Let's correct it:
mol = (35 mL) (0.1 mol/L) / (1000 mL) (0.20) * 2
The factor of 2 in this equation comes from the stoichiometry of the reaction in Step 2. Looking at the balanced equation, we can see that for each Ca2+ ion, we need 2 moles of calcium salt (CaCl2) as the starting material to produce 1 mole of Ca(OH)2. Therefore, we multiply by 2 to account for this stoichiometry.
So, the correct equation for the calculation of moles of Ca2+ per gram of calcium salt is:
mol = (35 mL) (0.1 mol/L) / (1000 mL) (0.20) * 2
I hope this explanation helps you understand the stoichiometry in this problem!