Chemistry, 14.07.2019 00:00 ayoismeisalex

You measured out 1.212g caco3 to create the calcium supplement. how many ml of 1m hcl would you need to completely dissolve the caco3?

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Answer from: daedae40

24.24 mL

Explanation: -

Mass of CaCO₃ = 1.212 g

Molar mass of CaCO₃ = 40 x 1 +12 x 1 + 16 x 3 = 100 g / mol

Number of moles of CaCO₃ = $\frac{1.212 g}{100 g/ mol}$

= 0.01212 mol

The balanced chemical equation for the reaction that occurs when CaCO₃ is added to HCl is

CaCO₃ + 2HCl --> CaCl₂ + H₂O + CO₂

From the equation we see

1 mol of CaCO₃ reacts with 2 mol of HCl

0.01212 mol of CaCO₃ react with $\frac{2 mol HCl x 0.01212 mol CaCO₃}{1 mol CaCO₃}$

= 0.02424 mol of HCl

Strength of HCl = 1 M

Volume of HCl required = Number of moles of HCl / strength of HCl

= 0.02424 mol / 1 M

= 0.02424 L

= 0.02424 x 1000 mL

= 24.24 mL

Answer from: Quest

alchemy hadn't scientific methods of study, appeal to mysticism and fantasy, the knowledge of chemistry and physics were modest at that time.

alchemy was - but as new rules were discovered it either changed into chemistry or it became "unscientific".

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³

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You measured out 1.212g caco3 to create the calcium supplement. how many ml of 1m hcl would you need...

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