1.64g of a mixture of cac03 and mhco3 was dissolved in 50ml of 0.8M hcl. The excess of acid required 16ml of 0.25M naoh for neutralization. Calculate the percentage of cac03 and mgc03 in the sample
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16ml of .25M NaOH = .004 moles NaOH
So, .004 moles HCl was needed to neutralize it
50ml of .8M HCl = .04 moles HCl
That means that .036 moles HCl reacted with the CO3 stuff
.018 moles of CO3 stuff reacted with the HCl.
CaCO3 has a mol wt of 100.09
MgCO3 has a mol wt of 84.31
x/100.09 + (1.64-x)84.31 = .018
x = .776g CaCO3
.776g CaCO3 = .00775 moles
.018-.00775 = .0102 moles MgCO3
You don't say whether you want the %ages in terms of moles or weight, so I'll leave that calculation to you.
I can help you with that! To calculate the percentage of CaCO3 and MgCO3 in the sample, we will use the concept of stoichiometry.
Let's break down the steps to solve this problem:
Step 1: Calculate the moles of HCl used.
To calculate the moles of HCl used, we will use the molarity and volume of the acid. The formula to calculate moles is:
moles = molarity x volume (in liters)
Given:
Molarity of HCl = 0.8 M
Volume of HCl = 50 mL (0.05 L)
Moles of HCl = 0.8 M x 0.05 L = 0.04 moles
Step 2: Calculate the moles of excess HCl.
To calculate the moles of excess HCl, we need to subtract the moles of NaOH used for neutralization. The moles of excess HCl can be calculated using the balanced chemical equation between HCl and NaOH.
The balanced chemical equation is:
HCl + NaOH -> NaCl + H2O
From the equation, we can see that 1 mole of HCl reacts with 1 mole of NaOH.
Given:
Molarity of NaOH = 0.25 M
Volume of NaOH = 16 mL (0.016 L)
Moles of NaOH = 0.25 M x 0.016 L = 0.004 moles (moles of HCl used)
Since the stoichiometry is 1:1, the moles of excess HCl is 0.04 moles - 0.004 moles = 0.036 moles.
Step 3: Calculate the moles of CaCO3 and MgCO3 in the mixture.
To calculate the moles of CaCO3 and MgCO3 in the mixture, we need to use the moles of excess HCl that reacted with the carbonates.
From the balanced chemical equation between HCl and CaCO3, we know that 1 mole of HCl reacts with 1 mole of CaCO3. Similarly, from the equation between HCl and MgCO3, 1 mole of HCl reacts with 1 mole of MgCO3.
Therefore, the moles of CaCO3 and MgCO3 in the mixture are both equal to 0.036 moles.
Step 4: Calculate the mass of CaCO3 and MgCO3.
To calculate the mass of CaCO3 and MgCO3 in the mixture, we need to use the molar mass of each compound.
The molar mass of CaCO3 = (1 x atomic mass of Ca) + (1 x atomic mass of C) + (3 x atomic mass of O)
The molar mass of MgCO3 = (1 x atomic mass of Mg) + (1 x atomic mass of C) + (3 x atomic mass of O)
After calculating the molar masses, we can multiply them by the moles of CaCO3 and MgCO3 calculated in Step 3 to get the masses.
Step 5: Calculate the percentage of CaCO3 and MgCO3.
To calculate the percentage, we need to divide the mass of each carbonate by the total mass of the mixture and multiply by 100.
Finally, the percentage of CaCO3 and MgCO3 in the sample can be determined.