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Tems11 [23]
2 years ago
7

Find the number of Fe2O3 moles needed to release 12 moles of CO2 . ​

Chemistry
1 answer:
miv72 [106K]2 years ago
3 0

Answer:

Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)

Explanation:

You might be interested in
A 50 L cylinder is filled with argon gas to a pressure of 10130.0 kPa at 300°C. How many moles of argon gas does the cylinder co
KengaRu [80]
In this problem, we need to use the ideal gas law. The following is the formula used in ideal gas law: PV = nRT, where n refers to the moles and R is the gas constant.

Given 
P = 10130.0 kPa 
V = 50 L
T = 300 degree celcius + 273.15 = 573.15 K
R = 8.314 L. kPa/K.mol

Solution
To get the moles which represent the "n" in the formula, we need to rearrange the equation.

PV = nRT                      PV
----    ------    --->    n = --------
 RT     RT                       RT

          10130.0 kPa  x 50 L
n= ---------------------------------------------
       8.314 L. kPa/K.mol  x 573.15 K
             506,500 
  =  ----------------------------
         4,765.17  mol K

=106.29 mol Ar

So the moles of argon gas is 106.29 moles 
8 0
3 years ago
Read 2 more answers
A chemist combines ethane and chlorine with the goal of producing chloroethane which other product could be reasonably expected
lesya692 [45]
<span>So when the chemist combines Ethane (CH3CH3) and Chlorine (Cl2) with the intention of producing Chloroethane (CH3CH2Cl), the other product that's formed in this reaction is 1,2-dichloroethane (ClCH2CH2Cl) also called as Ethylene dichloride with molecular weight of 98.954 g/mol. This is a colorless oily flammable substance that weighs heaver when vaporized.</span>
5 0
3 years ago
Read 2 more answers
When mixed, solutions of silver nitrate, AgNO3, and sodium sulfate, Na2SO4, form a precipitate of silver sulfate, Ag2SO4. The ba
Maslowich

Answer:

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

Explanation:

The general schemefor a reaction is given as;

Reactants --> Products

In this question, the reactants are AgNO3 and Na2SO4. The product is Ag2SO4.

The equation is given as;

AgNO3 + Na2SO4 --> Ag2SO4

The other poduct formed in this reaction is NaNO3.

The full reaction is given as;

AgNO3 + Na2SO4 --> Ag2SO4 + NaNO3

The above reaction is not balanced because there are unequal number of atoms of the elements on both sides of the reaction.

The balanced equation is given as;

2 AgNO3 + Na2SO4 → Ag2SO4 + 2 NaNO3

In this equation, there are equal number of moles of the atoms on both sides.

3 0
3 years ago
Which of the following possess the greatest concentration of hydroxide ions?
jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

11=-\log [OH^-]

[OH^-]=1.0\times 10^{-11}M

Thus, the OH^- concentration is, 1.0\times 10^{-11}M

<u>(b) a solution of 0.10 M NH_3</u>

As we know that 1 mole of NH_3 is a weak base. So, in a solution it will not dissociates completely.

So, the OH^- concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12=-\log [OH^-]

[OH^-]=1.0\times 10^{-12}M

Thus, the OH^- concentration is, 1.0\times 10^{-12}M

<u>(d) a solution of 0.10 M NaOH</u>

As we know that NaOH is a strong base. So, it dissociates to give Na^+ ion and OH^- ion.

So, 0.10 M of NaOH in a solution dissociates to give 0.10 M of Na^+ ion and 0.10 M of OH^- ion.

Thus, the OH^- concentration is, 0.10 M

<u>(e) a 1\times 10^{-4}M solution of HNO_2</u>

As we know that 1 mole of HNO_2 in a solution dissociates to give 1 mole of H^+ ion and 1 mole of NO_2^- ion.

So, 1\times 10^{-4}M of HNO_2 in a solution dissociates to give 1\times 10^{-4}M of H^+ ion and 1\times 10^{-4}M of NO_2^- ion.

The concentration of H^+ ion is 1\times 10^{-4}M

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (1.0\times 10^{-4})

pH=4

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

10=-\log [OH^-]

[OH^-]=1.0\times 10^{-10}M

Thus, the OH^- concentration is, 1.0\times 10^{-10}M

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0
3 years ago
68.3 grams of sodium hydroxide reacts with 78.3 grams of magnesium nitrate. ____ grams of magnesium hydroxide will form from thi
Vera_Pavlovna [14]

Answer:

30.8 grams of magnesium hydroxide will form from this reaction, and magnesium nitrate is the limiting reagent.

Explanation:

The reaction that takes place is:

  • 2NaOH + Mg(NO₃)₂ → 2NaNO₃ + Mg(OH)₂

Now we <u>convert the given masses of reactants to moles</u>, using their respective <em>molar masses</em>:

  • 68.3 g NaOH ÷ 40 g/mol = 1.71 mol NaOH
  • 78.3 g Mg(NO₃)₂ ÷ 148.3 g/mol = 0.528 mol Mg(NO₃)₂

0.528 moles of Mg(NO₃)₂ would react completely with (0.528 * 2) 1.056 moles of NaOH. There are more than enough NaOH moles, so NaOH is the reagent in excess and <em>Mg(NO₃)₂ is the limiting reagent.</em>

Now we <u>calculate how many Mg(OH)₂ are produced</u>, using the <em>moles of the limiting reagent</em>:

  • 0.528 mol Mg(NO₃)₂ * \frac{1molMg(OH)_2}{1molMg(NO_3)_2} = 0.528 mol Mg(OH)₂

Finally we convert Mg(OH)₂ moles to grams:

  • 0.528 mol Mg(OH)₂ * 58.32 g/mol = 30.8 g
7 0
2 years ago
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