Ask question

Ask question

All categories

3 years ago

11

A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti

on takes place: 2HCl(g) + Br2(g)2HBr(g) + Cl2(g) Calculate the volume of the sample after the reaction takes place, assuming that the temperature and the pressure remain constant.

1 answer:

Furkat [3]3 years ago

7 0

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

$2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)$

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

$n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2$

In such a way, the yielded moles of hydrobromic acid and chlorine are:

$n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2$

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

You might be interested in

1.) 3.4 moles Magnesium are mixed with 5.6 moles of Hydrochloric Acid. How many moles of Hydrogen gas are produced?

love history [14]

Answer:

1. 2.8 moles of H₂

2. 7.38 moles of CO₂

3. 5.3 moles of O₂

4. 7.4 moles of KNO₃

Explanation:

Here are the steps to doing this:

1. Write the chemical equation of each reaction.

2. Balance the equation.

3. Find out the ratio between reactant and product

4. Determine the actual yield of your reactants.

5. The amount of product produced is determined by how much product the limiting reactant produces.

Let's do this!

1. Given: 3.4 moles of Magnesium(Mg) and 5.6 moles of Hydrochloric acid (HCl)

<u>Equation:</u>

Mg + <u>2</u>HCl → MgCl₂ + H₂

<u>Reactant to Product ratio</u>

1 mole of Mg produces 1 mole of H₂ $\dfrac{1moleof Mg}{1mole of H_{2}}$

2 moles of HCl produces 1 mole of H₂ $\dfrac{2molesofHCl}{1mole of H_{2}}$

<u>Determine actual yield of reactants</u>

$3.4moles of Mg\times\dfrac{1moleofH_2}{1moleofMg}=3.4molesofH_{2}\\\\5.6moles ofHCl\times\dfrac{1moleofH_2}{2moleofHCl}=2.8molesofH_{2}$

Since 5.6 moles of HCl can only produce 2.8 moles of H₂, before it is used up, then this means that that is all the product this reaction can produce.

2. Given: 3.4 moles of C₃H₈ and 12.3 moles of oxygen gas (O₂)

<u>Equation:</u>

C₃H₈ + <u>5</u>O₂ → <u>3</u>CO₂ + <u>4</u>H₂O

<u>Reactant to Product ratio</u>

1 mole of C₃H₈ produces 3 moles of CO₂ $\dfrac{1moleofC_{3}H_{8}}{3molesofCO_{2}}$

5 moles of O₂ produces 3 moles of CO₂ $\dfrac{5molesofO_{2}}{3moleofCO_{2}}$

<u>Determine actual yield of reactants</u>

$3.4molesofC_{3}H_{8}\times\dfrac{3molesofCO_{2}}{1moleofC_{3}H_{8}}=10.2molesofH_{2}$

$12.3molesofO_{2}\times\dfrac{3molesofCO_{2}}{5molesofO_{2}}=7.38molesofCO_{2}$

The answer is then 7.38 moles of CO₂

**3. 5.3 moles of H₂O

This one is a little bit different. It is asking how much of a reactant is needed to produce the amount of product given. For this, just write a balanced equation for the reaction and get the ratio of reactant to product and solve for the actual yield. Since it is only asking for oxygen gas, you just need to do that one.

<u>Equation:</u>

CH₄ + <u>2</u>O₂ → CO₂ + <u>2</u>H₂O

<u>Reactant to Product ratio</u>

$\dfrac{2molesofO_{2}}{2molesofH_{2}O}=\[tex]7.88molesofKI\times\dfrac{1moleofKNO_{3}}{1moleofKI}=7.88molesofKNO_{3}$

<u>Actual yield:</u>

$5.3molesofH_{2}O\times\dfrac{1moleofO_{2}}{1moleofH_{2}O}=5.3molesofO_{2}$

The answer is 5.3 moles of O₂.

4. 3.7 moles of Lead (II) Nitrate (Pb(NO₃)₂) and 7.8 moles of Potassium Iodide (KI)

<u>Equation:</u>

Pb(NO₃)₂ + <u>2</u>KI → PbI₂ + <u>2</u>KNO₃

<u>Reactant to Product ratio</u>

$\dfrac{2molesofKI}{2molesofKNO_{3}}=\dfrac{1moleofKI}{1moleofKNO_{3}}$

$\dfrac{1molesofPb(NO_{3})_{2}}{2molesofKNO_{3}}$

<u>Actual yield:</u>

$3.7molesofPb(NO_{3})_{2}\times\dfrac{2molesofKNO_{3}}{1moleofPb(NO_{3})_{2}}=7.4molesofKNO_{3}$

The answer is 7.4 moles of KNO₃.

3 0

3 years ago

Which term goes hand in hand with biodiversity?

Romashka-Z-Leto [24]

I think it’s B and why is everyone putting a bunch of links?

7 0

3 years ago

A 12.0g sample of a substance requires 50.9 J to raise its temperature by 6.00 c. What is it’s specific heat

horsena [70]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's find the specific heat of sample using given formula :

$\qquad \tt \dashrightarrow \:Q = m \: s \triangle T$

where,

Q = heat transfer

m = mass

S = specific heat

$\tt \triangle T$ = change in temperature

$\qquad \tt \dashrightarrow \:50.9 = 12 \times \: s \times 6$

$\qquad \tt \dashrightarrow \:s = \dfrac{50.9}{72}$

$\qquad \tt \dashrightarrow \:s = 0.707 \:$

Therefore, specific heat of that substance is 0.707 J/g°C

7 0

2 years ago

Ammonia is a?<br> Base<br> Acid<br> Ketone<br> Ester

BaLLatris [955]

Answer:

it is a base

Explanation:

8 0

2 years ago

A hydrogen ion, H+, is the same as a.... <br> PLEASEEE

sdas [7]

It is the same as a proton. That is why, if hydrogen was a molecule not an ion, one electron is needed to 'cancel' out the proton.

5 0

2 years ago