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klemol [59]
3 years ago
9

Carbon disulfide is prepared industrially by reacting carbon with sulfur dioxide according to the above equation. If 70.8 g of c

arbon react, how many moles of CS2 are produced? Show all work, include all units, and box your answer.
Chemistry
1 answer:
Rus_ich [418]3 years ago
8 0

Answer:

1.18 moles of CS₂ are produced by the reaction.

Explanation:

We present the reaction:

5C + 2SO₂  →  CS₂  +  4CO

5 moles of carbon react to 2 moles of sulfur dioxide in order to produce 1 mol of carbon disulfide and 4 moles of carbon monoxide.

As we do not have data from the SO₂, we assume this as the excess reagent. We convert the mass of carbon to moles:

70.8 g / 12 g/mol = 5.9 moles

Ratio is 5:1, so 5 moles of carbon react to produce 1 mol of CS₂

Then, 5.9 moles will produce (5.9 . 1) / 5 = 1.18 moles

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If copper carbonate system is heated for too short a time, what will you notice for mass percent of copper be too large or too s
sammy [17]

Answer:

The mass percent of copper as element is the same.

Explanation:

First of all we need the reaction that is presented below:

CuCO_{3 (s)} +Heat → CuO_{(s)} + CO_{2 (g)}

The mass percent of copper (Cu) as element is the same because of during the reaction the element only transform its nature from copper carbonate (CuCO_{3 (s)}) to copper oxide (CuO_{(s)}), the latter is a solid and will remain in the system.

On the other hand, you will note that the global percentage mass will be small because of the reaction produce (CO_{2 (g)}) that is a gas and this one will escape for the system.  

Have a great day!

3 0
3 years ago
PLEASE ANSWER 30 POINTS !!
Alexeev081 [22]
Can u explain it more plz.
3 0
3 years ago
Read 2 more answers
What is the concentration of hydronium ion in a 0.121 M HCl solution?
kogti [31]

Answer:

C) 0.121 M

Explanation:

HCl + H₂O = H₃O⁺ + OH⁻

.121M              .121M

HCl is a strong acid . It will dissociate almost 100 % so the concentration of acid and hydronium ion formed will be equal . It is to be noted that hydronium ion is formed due to association of H⁺ and H₂O . H⁺ is formed due to ionisation of HCl .

So concentrtion of hydronium ion ( H₃O⁺ ) will be .121 M.

7 0
3 years ago
Match: Substance ....... Definition Lewis acid A. Provides H+ in water Bronsted-Lowry base B. Provides OH - in water Arrhenius b
Igoryamba

Answer: F.    Electron pair acceptor

Explanation:

A Lewis acid can be properly defined as any substance such as  H+ (hygrogen ion) that can accept a pair of electron.

While a Lewis base is any substance such as (OH-) that can donate a pair of electron.

In the neutralization reaction between an acid ( H+ ) and a base (OH-). Hydrogen ion (H+ ) is the Lewis acid because it accepts an electron pair from (OH-).

Other examples of Lewis acid are;  Mg2+, K+

3 0
3 years ago
If 16.00 g of O₂ reacts with 80.00 g NO, how many the excess reactant are left over? (enter only the value, round to whole numbe
pishuonlain [190]

Answer:

50

Explanation:

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

1. Gather all the information in one place with molar masses above the formulas and masses below them.  

Mᵣ:           30.01     32.00   46.01

               2NO   +   O₂ ⟶ 2NO₂

Mass/g:  80.00     16.00

2. Calculate the moles of each reactant  

\text{moles of NO} = \text{80.00 g NO} \times \dfrac{\text{1 mol NO}}{\text{30.01 g NO}} = \text{2.666 mol NO}\\\\\text{moles of O}_{2} = \text{16.00 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.5000 mol O}_{2}

3. Calculate the moles of NO₂ we can obtain from each reactant

From NO:

The molar ratio is 2 mol NO₂:2 mol NO

\text{Moles of NO}_{2} = \text{2.333 mol NO} \times \dfrac{\text{2 mol NO}_{2}}{\text{2 mol NO}} = \text{2.333 mol NO}_{2}

From O₂:

The molar ratio is 2 mol NO₂:1 mol O₂

\text{Moles of NO}_{2} =  \text{0.5000 mol O}_{2}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol Cl}_{2}} = \text{1.000 mol NO}_{2}

4. Identify the limiting and excess reactants

The limiting reactant is O₂ because it gives the smaller amount of NO₂.

The excess reactant is NO.

5. Mass of excess reactant

(a) Moles of NO reacted

The molar ratio is 2 mol NO:1 mol O₂

\text{Moles reacted} = \text{0.500 mol O}_{2} \times \dfrac{\text{2 mol NO}}{\text{1 mol O}_{2}} = \text{1.000 mol NO}

(b) Mass of NO reacted

\text{Mass reacted} = \text{1.000 mol NO} \times \dfrac{\text{30.01 g NO}}{\text{1 mol NO}} = \text{30.01 g NO}

(c) Mass of NO remaining

Mass remaining = original mass – mass reacted = (80.00 - 30.01) g = 50 g NO

5 0
3 years ago
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