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Scilla [17]
3 years ago
11

If 294 grams of FeS2 is allowed ti react with 176 grams of O2 according to the following equation how many grams of Fe2O3 are pr

oduced?
FeS2 + O2 = Fe2O3 + SO2
Chemistry
2 answers:
podryga [215]3 years ago
6 0
 <span>You must balance your equation correctly.
Here is your answer: 

294gFeS2 x 1molFeS2/119.99 x 11mols O2/4mols FeS2--> 6.738mol O2 

176gO2 x 1mol O2/32gO2 x 4mols FeS2/11mol FeS2--> 2mols FeS2 
Now choose the molecule with the lowest amount (Limiting Reagent) 

2molsFeS2 x 2molsFe2O3/4molsFeS2 x 159.7g
159.7g Fe2O3 grams produced.</span>
bija089 [108]3 years ago
3 0

Answer:

Amount of Fe2O3 produced is 197 g

Explanation:

Step 1: Deduce the limiting reactant

Mass of FeS2 = 294 g

Molar Mass of FeS2 = 120 g/mol

Moles\ FeS2 = \frac{Mass}{Molar Mass} =\frac{294}{120} =2.45 moles

Mass of O2 = 176 g

Molar mass of O2 = 32 g/mol

Moles\ O2 = \frac{Mass}{Molar Mass} =\frac{176}{32} =5.5 moles

Since moles of FeS2 < O2, then FeS2 is the limiting reactant which will dictate the amount of product formed

Step 2: Calculate the mass of Fe2O3 produced

The balanced equation is:

4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Based on the reaction stoichiometry:

4 moles of FeS2 produced 2 moles of Fe2O3

Therefore, 2.45 moles of FeS2 will produce 1.23 moles Fe2O3

Molar mass of Fe2O3 = 160 g/mol

Mass\ of\ Fe2O3 = moles * molar\ mass = 1.23 * 160 = 196.8 g

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AlekseyPX

Agriculture, space exploration, and also for medical purposes.

Hope I helped :)

6 0
3 years ago
If equal volumes of 0.1 M HCl and 0.2 M TRIS (base form) are mixed together. The pKa of TRIS is 8.30. Which of the following sta
blondinia [14]

Answer:

option D is correct

D. This solution is a good buffer.

Explanation:

TRIS (HOCH_{2})_{3}CNH_{2}

if TRIS is react with HCL it will form salt

(HOCH_{2})_{3}CNH_{2} + HCL ⇆   (HOCH_{2})_{3}NH_{3}CL

Let the reference volume is 100

Mole of TRIS is =  100 × 0.2 = 20

Mole of HCL is =  100 × 0.1 = 10

In the reaction all of the HCL will Consumed,10 moles of the salt will form

and 10 mole of TRIS will left

hence , Final product will be salt +TRIS(9 base)

H = Pk_{a} + log (base/ acid)

8.3 + log(10/10)

8.3

6 0
3 years ago
What is the formula of haso4​
ladessa [460]

Explanation:

The structure of Ferrarrisite Ca5(HAs O4)2(AsO4)2

8 0
3 years ago
Ca(OH)2 (s) precipitates when a 1.0 g sample of CaC2(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g s
Nina [5.8K]

Answer:

D) Ca(OH)₂ will not precipitate because Q <  Ksp

Explanation:

Here we have first a chemical reaction in which Ca(OH)₂  is produced:

CaC₂(s)  + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂  is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product  constant for Ca(OH)₂  is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M  = 0.002 M  ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q=  [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is option D

8 0
3 years ago
How many kilograms of potassium iodide (ki) are needed to make 1.25 l of a 4.41 m ki solution?
Lapatulllka [165]
C = n/V
n = C×V
n = 4,41M × 1,25L
n = 5,5125 mol

mKI: 39+127 = 166 g/mol

1 mol --------- 166g
5,5125 mol --- X
X = 166×5,5125 = 915,075g KI

:)
7 0
3 years ago
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