Chemistry, 29.02.2020 00:11 reddmeans6

If you react 0.025 moles of 2-methylbromobenzene, 0.7 grams of Mg, and 10 grams of dry ice, what is the theoretical yield of 2-methylbenzoic acid in grams?

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Answer from: tatertott6876

$\large \boxed{\text{3.4 g}}$

Explanation:

1. Write the overall equation.

The Grignard reaction has several steps:

(a) CH₃C₆H₄Br + Mg ⟶ CH₃C₆H₄MgBr

(b) CH₃C₆H₄MgBr + CO₂ ⟶ CH₃C₆H₄COOMgBr

CH₃C₆H₄Br + Mg + CO₂ + H⁺ ⟶ CH₃C₆H₄COOH + Mg²⁺ + Br⁻

Now we can gather all the information in one place.

Mᵣ: 171.03 24.31 44.01 136.15

CH₃C₆H₄Br + Mg + CO₂ + H⁺ ⟶ CH₃C₆H₄COOH + Mg²⁺ + Br⁻

n/mol: 0.025

m/g: 0.7 10

Let's call CH₃C₆H₄Br "R" (reactant) and CH₃C₆H₄COOH "P" (product).

They have given us the amounts of three reactants and asked us to calculate the amount of product. This is a limiting reactant problem.

1. Calculate the moles of each reactant

(a) CH₃C₆H₄Br

n = 0.025 mol (given)

(b) Mg

$\text{Moles of Mg} = \text{0.7 g Mg} \times \dfrac{\text{1 mol Mg}}{\text{24.31 g Mg}} = \text{0.029 mol Mg}$

$\text{Moles of CO}_{2}= \text{10 g CO}_{2} \times \dfrac{\text{1 mol CO}_{2}}{\text{44.01 g CO}_{2}} = \text{0.23 mol CO}_{2}$

2. Calculate the moles of CH₃C₆H₄COOH you can obtain from each reactant

The molar ratios are all 1:1, so

(a) 0.025 mol R ⟶ 0.025 mol P

(b) 0.028 mol Mg ⟶ 0.029 mol P

R is the limiting reactant, because it gives the fewest moles of P.

3. Calculate the theoretical yield.

$\text{Mass of P} = \text{0.025 mol P} \times \dfrac{\text{136.15 g P}}{\text{1 mol P}} = \textbf{3.4 g P}\\\\\text{The theoretical yield of 2-methylbenzoic acid is $\large \boxed{\textbf{3.4 g}}$}$

Answer from: Quest

Ametal with a large number of free flowing electrons most likely has good conductivity

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

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If you react 0.025 moles of 2-methylbromobenzene, 0.7 grams of Mg, and 10 grams of dry ice, what is...

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