Chemistry, 18.01.2021 07:00 dakotaadkins1818

3. If 12.5 grams of copper are reacted with excess chlorine gas, then 25.4 grams of copper (II) chloride are produced. Calculate the theoretical yield and the percent
yield. Cu + Cl2 → CuCl2

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Answer from: adelet

Percent yield = 99.6%

Theoretical yield = 25.5 g

Explanation:

Given data:

Mass of copper = 12.5 g

Mass of copper(II) chloride produced = 25.4 g

Theoretical yield = ?

Percent yield = ?

Solution:

Chemical equation:

Cu + Cl₂ → CuCl₂

Number of moles of copper:

Number of moles = mass/molar mass

Number of moles = 12.5 g / 63.5 g/mol

Number of moles = 0.19 mol

Now we will compare the moles of copper and copper chloride.

Cu : CuCl₂

1 : 1

0.19 : 0.19

Mass of CuCl₂: Theoretical yield

Mass = number of moles × molar mass

Mass = 0.19 mol × 134.45 g/mol

Mass = 25.5 g

Percent yield:

Percent yield = (actual yield / theoretical yield )×100

Percent yield = ( 25.4 g/ 25.5 g) ×100

Percent yield = 99.6%

Answer from: Quest

What essay do you want us to write?

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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