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Tom [10]
3 years ago
11

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed

Chemistry
1 answer:
MaRussiya [10]3 years ago
3 0

The question is incomplete, here is the complete question:

Suppose the reaction is:

A+2B\rightarrow C

If 6 moles of A and 2 moles of B are reacted, what is the maximum number of moles of C that can be formed

<u>Answer:</u> The maximum amount of C produced will be 1 mole

<u>Explanation:</u>

We are given:

Moles of A = 6 moles

Moles of B = 2 moles

For the given chemical reaction:

A+2B\rightarrow C

By Stoichiometry of the reaction:

2 moles of B reacts with 1 mole of A

So, 2 moles of B will react with = \frac{1}{2}\times 2=1 mole of A

As, As, given amount of A is more than the required amount. So, it is considered as an excess reagent.

Thus, B is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of B produces 1 mole of C

So, 2 moles of given B will produce =\frac{1}{2}\times 2=1mol of C

Hence, the maximum amount of C produced will be 1 mole

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The value of the solubility product constant for Ag2CO3 is 8.5 × 10‒12 and that of Ag2CrO4 is 1.1 × 10‒12. From this data, what
Lena [83]

Answer:

B) 7.7

Explanation:

For the reaction    Ag2CO3(s) + CrO42‒(aq) → Ag2CrO4(s) + CO32‒(aq)

Kc = (CO₃²⁻) / (CrO₄²⁻)

and the Ksp given are

Ag₂CO₃    ⇒  2 Ag⁺(aq) + CO₃²⁻(aq)    Ksp₁ = (Ag⁺)²(CO₃²⁻)  

Ag₂CrO₄   ⇒  2 Ag⁺(aq)+ CrO₄²⁻(aq)   Ksp₂ = (Ag⁺)²(CrO₄²⁻)

Where (...) indicate concentrations M

Notice if we divide the expressions for Ksp we get:

Ksp₁/Ksp₂ = (CO₃²⁻)  / (CrO₄²⁻) = 8.5 x 10⁻¹² / 1.1 x 10⁻¹² = 7.7

which is the desired answer.

7 0
3 years ago
URGENT !! A substance has 55.80% carbon, 7.04% Hydrogen, and 37.16% Oxygen. What is it's empirical and molecular formula if it h
Ede4ka [16]
<h3><u>Answer;</u></h3>

Empirical formula = C₂H₃O

Molecular formula = C₁₄H₂₁O₇

<h3><u>Explanation</u>;</h3>

Empirical formula

Moles of;

Carbon = 55.8 /12 = 4.65 moles

Hydrogen = 7.04/ 1 = 7.04 moles

Oxygen  = 37.16/ 16 = 2.3225 moles

We then get the mole ratio;

4.65/2.3225 = 2.0

7.04/2.3225 = 3.0

2.3225/2.3225 = 1.0

Therefore;

The empirical formula = <u>C₂H₃O</u>

Molecular formula;

(C2H3O)n = 301.35 g

(12 ×2 + 3× 1 + 16×1)n = 301.35

43n = 301.35

  n = 7

Therefore;

Molecular formula = (C2H3O)7

                             <u> = C₁₄H₂₁O₇</u>

6 0
4 years ago
Average Molarity for HCl is .391
Ira Lisetskai [31]

Answer:

#1: 0.00144 mmolHCl/mg Sample

#2: 0.00155 mmolHCl/mg Sample

#3: 0.00153 mmolHCl/mg Sample

Explanation:

A antiacid (weak base) will react with the HCl thus:

Antiacid + HCl → Water + Salt.

In the titration of antiacid, the strong acid (HCl)  is added in excess, and you're titrating with NaOH moles of HCl that doesn't react.

Moles that react are the difference between mmoles of HCl - mmoles NaOH added (mmoles are Molarity×mL added). Thus:

Trial 1: 0.391M×14.00mL - 0.0962M×34.26mL = 2.178 mmoles HCl

Trial 2: 0.391M×14.00mL - 0.0962M×33.48mL = 2.253 mmoles HCl

Trial 3: 0.391M×14.00mL - 0.0962M×33.84mL = 2.219 mmoles HCl

The mass of tablet in mg in the 3 experiments is 1515mg, 1452mg and 1443mg.

Thus, mmoles HCl /mg OF SAMPLE<em> </em>for each trial is:

#1: 2.178mmol / 1515mg

#2: 2.253mmol / 1452mg

#3: 2.219mmol / 1443mg

<h3>#1: 0.00144 mmolHCl/mg Sample</h3><h3>#2: 0.00155 mmolHCl/mg Sample</h3><h3>#3: 0.00153 mmolHCl/mg Sample</h3>
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Answer:

The answers are A,B,C.

Explanation: Just got it right on Edge 2020

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Hydrogen is the first element listed on the periodic table and can be found in the top leftmost position. Oxygen is the eighth element on the periodic able and is located in the second row at the top right.

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2 years ago
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