Chemistry, 11.04.2020 00:53 kaylijocombs

The active component in one type of calcium dietary supplement is calcium carbonate. A 1.2450 g tablet of the supplement is added to 65.00 mL of 0.4984 M HCl(aq) and allowed to react. After completion of the reaction, the excess HCl(aq) requires 37.15 mL of 0.2065 M NaOH(aq) for its titration to the equivalence point. What is the calcium content of the tablet, expressed in milligrams of Ca?

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Answer from: andrewweltz421

The calcium content of the tablet is 486 mg.

The reaction between the HCl and the supplement occurs as follows;

CaCO3(s) + 2HCl(aq) > CaCl2(aq) + CO2(g) + H2O(l)

Total amount of HCl reacted = 65/100 L × 0.4984 M = 0.032 moles

From the reaction;

HCl(aq) + NaOH(aq) ---> NaCl (aq) + H2O(l)

Number of moles of NaOH = 37.15/1000 × 0.2065 M = 0.0077 moles

Since the reaction is 1:1, 0.0077 M of excess HCl is present

Amount of HCl reacted = 0.032 moles - 0.0077 moles = 0.0243 moles

1 mole of calcium carbonate reacts with 2 moles of HCl

x moles of calcium carbonate reacts with 0.0243 moles of HCl

x = 1 mole × 0.0243 moles/2 moles

x = 0.01215 moles

Mass of CaCO3 reacted = 0.01215 moles × 100 g/mol = 1.215 g

If 100 g of CaCO3 contains 40 g of Ca

1.1215 g of CaCO3 contains 1.215 g × 40 g/100 g = 0.486 g or 486 mg

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Answer from: srice6

$\large \boxed{\text{495.5 mg }}$

Explanation:

This is an example of back titration.

You add a known excess of HCl to the CaCO₃. Some of it reacts with the CaCO₃.

You titrate the left-over HCl.

From the difference you can calculate the amount of CaCO₃.

100.09

CaCO₃ + 2HCl ⟶ CaCl₂ + H₂CO₃

m/g: 1.2450

V/mL: 65.00

c/mol·L⁻¹: 0.4984

NaOH + HCl ⟶ NaCl + H₂O

V/mL: 37.15

c/mol·L⁻¹: 0.2065

1. Total moles of HCl

$\text{Moles of HCl}= \text{65.00 mL HCl} \times \dfrac{\text{0.4984 mmol HCl}}{\text{1 mL HCl}} = \text{32.40 mmol HCl}$

2. Excess moles of HCl

$\text{Moles of HCl} = \text{37.15 mL NaOH} \times \dfrac {\text{0.2065 mmol NaOH}}{\text{1 mL NaOH}} \times \dfrac{\text{1 mmol HCl}}{\text{1 mmol NaOH}}\\\\= \text{7.671 mmol HCl}$

3. Moles of HCl that reacted

n = (32.40 - 7.671) mmol HCl = 24.72 mmol HCl

4. Moles of CaCO₃

$\text{Moles of CaCO}_{3} = \text{24.72 mmol HCl} \times \dfrac{\text{1 mmol CaCO}_{3}}{\text{2 mmol HCl}} = \text{12.36 mmol CaCO}_{3}$

5. Mass of Ca

$\text{Mass of Ca} = \text{12.36 mmol CaCO}_{3} \times \dfrac{\text{1 mmol Ca}}{\text{ 1 mmol CaCO}_{3}} \times \dfrac{\text{40.08 mg Ca}}{\text{1 mmol Ca}}\\\\= \textbf{495.5 mg Ca}\\\\\text{The calcium content of the tablet is $\large \boxed{\textbf{495.5 mg }}$}$

Answer from: Quest

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.

Answer from: Quest

answer: they produce more slow than unicellular organisms.

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