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3 years ago

What is the percent yield for a reaction if the theoretical yield of Fe2(SO4) is 33.14 g and the

Chemistry

1 answer:

Triss [41]3 years ago

4 0

Answer:

% yield = 55.82%

Explanation:

Theoretical yield = 33.14g

Actual yield = 18.5g

Percentage yield = (actual yield / theoretical yield) * 100

%yield = (18.5 / 33.14) * 100

% yield = 0.5582 * 100

% yield = 55.82%

The percentage yield of Fe₂So₄ is 55.82%

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A scientific ________ must have a control, so that the variables that could affect the out come is reduced. A:experiment B:concl

Lostsunrise [7]

Answer:

A.experiment

Explanation:

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4 0

3 years ago

Determine the value of the equilibrium constant, KgoalKgoalK_goal, for the reaction CO2(g)⇌C(s)+O2(g)CO2(g)⇌C(s)+O2(g), Kgoal=?

NISA [10]

Answer:

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

Explanation:

$CO_2(g)\rightleftharpoons C(s)+O_2(g)$

$K_{goal}=?$

$K_{goal}=\frac{[C][O_2]}{[CO_2]}$

$2CO_2(g)+2H_2O(l)\rightleftharpoons CH_3COOH(l)+2O_2(g)$ ..[1]

$K_1=\frac{[CH_3COOH][O_2]^2}{[CO_2]^2[H_2O]^2}$

$2H_2(g)+O2(g)\rightleftharpoons 2H_2O(l)$ ..[2]

$K_2=\frac{[H_2O]^2}{[H_2]^2[O_2]}$

$CH_3COOH(l)\rightleftharpoons 2C(s)+2H_2(g)+O_2(g)$ ..[3]

$K_3=\frac{[C]^2[H_2]^2[O_2]}{[CH_3COOH]}$

[1] + [2] + [3]

$2CO_2(g)\rightleftharpoons 2C(s)+2O_2(g)$

( on adding the equilibrium constant will get multiplied with each other)

$K=K_1\times K_2\times K_3$

$K=5.40\times 10^{-16}\times 1.06\times 10^{10}\times 2.68\times 10^{-9}$

$K=1.53\times 10^{-14}$

$K=\frac{[C]^2[O_2]^2}{[CO_2]^2}$

On comparing the K and $K_{goal}$ :

$K^2=K_{goal}$

$K_{goal}=\sqrt{K}=\sqrt{1.53\times 10^{-14}}=1.24\times 10^{-7}$

The value of the equilibrium constant for reaction asked is $1.24\times 10^{-7}$ .

4 0

3 years ago

Hydrogen sulfide decomposes according to the following reaction: 2H2S(g) ⇋ 2H2(g) + S2(g) Kc=9.30x10-8 at 700.°C.If 0.45 mol of

Natalija [7]

Answer:

[H₂] = 1.61x10⁻³ M

Explanation:

2H₂S(g) ⇋ 2H₂(g) + S₂(g)

Kc = 9.30x10⁻⁸ = $\frac{[H_{2}]^2[S_{2}]}{[H_{2}S]^2}$

First we calculate the initial concentration:

0.45 molH₂S / 3.0L = 0.15 M

The concentrations at equilibrium would be:

[H₂S] = 0.15 - 2x

[H₂] = 2x

[S₂] = x

We put the data in the Kc expression and solve for x:

$\frac{(2x^2) * x}{(0.15-2x)^2}=9.30x10^{-8}$

$\frac{4x^3}{0.0225-4x^2}=9.30*10^{-8}$

We make a simplification because x<<< 0.0225:

$\frac{4x^3}{0.0225} =9.30*10^{-8}$

x = 8.058x10⁻⁴

[H₂] = 2*x = 1.61x10⁻³ M

5 0

3 years ago

Read 2 more answers

9.63 Km is the equivalent to 9,630 m. True. or False

Karo-lina-s [1.5K]

Answer:

9.63 Km is the equivalent to 9,630 m.

9.63 Km is the equivalent to 9,630 m. TRUE

7 0

3 years ago

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Mama L [17]

1. Friction
2.conduction
3.induction
4. Conduction

8 0

2 years ago