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3 years ago

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15.00 grams of Chromium react with 15.00 grams of hydrobromic acid. Calculate the theoretical yield of the reaction. At STP what

volume would the gas occupy? If the actual yield is 15.35g chromium(III) bromide, calculate the percent yield of the reaction.

1 answer:

s2008m [1.1K]3 years ago

8 0

Answer:

(a) 18.03 g

(b) 2.105 L

(c) 85.15 %

Step-by-step explanation:

We have the masses of two reactants, so this is a<em> limiting reactant problem. </em>

We know that we will need a balanced equation with masses, moles, and molar masses of the compounds involved.

<em>Step 1</em>. <em>Gather all the information</em> in one place with molar masses above the formulas and masses below them.

M_r: 52.00 80.91 291.71

2Cr + 6HBr ⟶ 2CrBr₃ + 3H₂

Mass/g: 15.00 15.00

<em>Step 2</em>. Calculate the <em>moles of each reactant</em>

Moles of Cr = 15.00 × 1/52.00

Moles of Cr = 0.2885 mol Cr

Moles of HBr = 15.00 × 1/80.91

Moles of HBr = 0.1854 mol HBr ×

<em>Step 3</em>. Identify the<em> limiting reactant</em>

Calculate the moles of CrCl₃ we can obtain from each reactant.

<em>From Cr</em>:

The molar ratio of CrBr₃:Cr is 2 mol CrBr₃:2 mol Cr

Moles of CrBr₃ = 0.2885 × 2/2

Moles of CrBr₃ = 0.2885 mol CrCl₃

<em>From HBr: </em>

The molar ratio of CrBr₃:HBr is 2 mol CrBr₃:6 mol HBr.

Moles of CrBr₃ = 0.1854 × 2/6

Moles of CrBr₃ = 0.061 80 mol CrBr₃

The limiting reactant is HBr because it gives the smaller amount of CrBr₃.

<em>Step 4</em>. Calculate the <em>theoretical yields</em> of CrBr₃ and H₂.

Theoretical yield of CrBr₃ = 0.061 80 × 291.71/1

Theoretical yield of CrBr₃ = 18.03 g CrCl₃

The molar ratio is 3 mol H₂:6 mol HBr

Theoretical yield of H₂ = 0.1854 × 3/6

Theoretical yield of H₂ = 0.092 70 mol H₂

<em>Step 5</em>. Calculate the <em>volume of H₂</em> at STP

STP is 1 bar and 0 °C.

The molar volume of a gas at STP is 22.71 L.

Volume = 0.092 70 × 22.71/1

Volume = 2.105 L

<em>Step 6</em>. Calculate the <em>percent yield </em>

% Yield = actual yield/theoretical yield × 100 %

Actual yield = 15.35 g

% yield = 15.35/18.03 × 100

% yield = <em>85.15 % </em>

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Radiation is energy that comes from a source and travels through space at the speed of light.

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The equilibrium constant for the reaction

FinnZ [79.3K]

Answer: The concentrations of $Cl_2$ at equilibrium is 0.023 M

Explanation:

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Volume of solution = 1 L

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The given balanced equilibrium reaction is,

$COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)$

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The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}$

Now put all the given values in this expression, we get :

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x = 0.023 M

Thus, the concentrations of $Cl_2$ at equilibrium is 0.023 M

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3 years ago

Henry divides 1.060g by 1.0mlbto find the density of his water sample. How many significant figures should he include in the den

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3 years ago

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, <u><em>the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.</em></u>

5 0

3 years ago

A frictionless piston cylinder device is subjected to 1.013 bar external pressure. The piston mass is 200 kg, it has an area of

Bad White [126]

Answer:

a) $T_{2} = 360.955\,K$ , $P_{2} = 138569.171\,Pa\,(1.386\,bar)$ , b) $T_{2} = 347.348\,K$ , $V_{2} = 0.14\,m^{3}$

Explanation:

a) The ideal gas is experimenting an isocoric process and the following relationship is used:

$\frac{T_{1}}{P_{1}} = \frac{T_{2}}{P_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{v}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{v}}$

The number of moles of the ideal gas is:

$n = \frac{P_{1}\cdot V_{1}}{R_{u}\cdot T_{1}}$

$n = \frac{\left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}} \right)\cdot (0.12\,m^{3})}{(8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K} )\cdot (298\,K)}$

$n = 5.541\,mol$

The final temperature is:

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (30.1\,\frac{J}{mol\cdot K} )}$

$T_{2} = 360.955\,K$

The final pressure is:

$P_{2} = \frac{T_{2}}{T_{1}}\cdot P_{1}$

$P_{2} = \frac{360.955\,K}{298\,K}\cdot \left(101,325\,Pa + \frac{(200\,kg)\cdot (9.807\,\frac{m}{s^{2}} )}{0.15\,m^{2}}\right)$

$P_{2} = 138569.171\,Pa\,(1.386\,bar)$

b) The ideal gas is experimenting an isobaric process and the following relationship is used:

$\frac{T_{1}}{V_{1}} = \frac{T_{2}}{V_{2}}$

Final temperature is cleared from this expression:

$Q = n\cdot \bar c_{p}\cdot (T_{2}-T_{1})$

$T_{2} = T_{1} + \frac{Q}{n\cdot \bar c_{p}}$

$T_{2} = 298\,K +\frac{10,500\,J}{(5.541\,mol)\cdot (38.4\,\frac{J}{mol\cdot K} )}$

$T_{2} = 347.348\,K$

The final volume is:

$V_{2} = \frac{T_{2}}{T_{1}}\cdot V_{1}$

$V_{2} = \frac{347.348\,K}{298\,K}\cdot (0.12\,m^{3})$

$V_{2} = 0.14\,m^{3}$

4 0

3 years ago