please don\'t rush i am looking for a good answer here we took the average of th

Question

please don't rush i am looking for a good answer

here we took the average of the best three

Calculate the mass, in grams, of solid NaOH pellets required to have a 1 L solution of NaOh, 0.1 M. Show the calculation on the report form. Weight the NaOH and record the exact mass of NaOH on the report form. Prepare the NaOH in the 1 L polypropylene bottle. Fill the 1 bottle with distilled H2O until almost full. The exact volume of water is not important. The concentration is approximately 0.1 M. Calculate the mass of H2C2O4-2H2O, in grams, required to make a 250 ml solution of H2C2O4, 0.15 M.

Explanation / Answer

1. The equivalence equation for this is:

M1V1 / n1 = M2V2 / n2

where the subscript 1 is for H2C2O4.2H2O and subscript 2 for NaOH.

We have

n2x M1V1 / n1 = 3.032 x 10-3 mol

V2 = 27.90 mL from table 1.1

M2 = 3.032 x 10-3 mol / 0.0279 L = 0.109 mol/L

2. 50 mL of 0.25 M HCl reacts with the tablets.

Excess HCl reacts with NaOH.

The equivalence equation for NaOH and HCl is

MHClVHCl = MNaOHVNaOH

VNaOH = 13.416 mL

VHCl = 50 mL

Using concentration of NaOH obtained above, we get

MHCl = 0.109 x 13.416 / 50 = 0.029 mol/L

Moles of HCl reacted = (0.243 - MHCl) x 0.05 L = 1.069 x 10-2 moles

2 moles of HCl react with 1 mole of CaCO3

So, 0.01069 moles of HCl react with = 0.01069 / 2 = 5.346 x 10-3 moles of CaCO3

Molecular mass of Calcium Carbonate = 40 + 12 + 48 = 100 g/mol

Mass of CaCO3 in each tablet = 5.346 x 10-3 x 100 = 0.535 g

For sample 1,

% CaCO3 = 100 x 0.535 / 1.324 = 40.378 %

For sample 2,

% CaCO3 = 100 x 0.535 / 1.325 = 40.347 %

For sample 3,

% CaCO3 = 100 x 0.535 / 1.306 = 40.934 %

3.a. CaCO3 + 2HCl -------> CaCl2 + H2O + CO2

The gas produced is Carbon dioxide.

b. For 1 mole of Calcium Carbonate we have 1 mole of Carbon dioxide produced

For 500 mg of Calcium Carbonate = 0.5 g / 100 g/mol = 5 x 10-3 mol, we shall have 5 x 10-3 moles of Carbon dioxide produced.

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