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RideAnS [48]
4 years ago
5

Balance the following equation and answer the question. ___CuCl + ___ H2S ------ ___ Cu2S + ___ HCl If 1.2 moles of CuCl is mixe

d with an excess of H2S gas, how many moles of Cu2S would be made? *
Chemistry
1 answer:
brilliants [131]4 years ago
3 0

Answer:

A. 2CuCl + H2S —> Cu2S + 2HCl

B. 0.6 mole of Cu2S

Explanation:

A. The balancing equation.

The equation can be balance as follow:

CuCl + H2S —> Cu2S + HCl

There are 2 atoms of Cu on the right side and 1 atom on the left. It can be balance by putting 2 in front of CuCl as shown below:

2CuCl + H2S —> Cu2S + HCl

There are 2 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 2 in front of HCl as shown below:

2CuCl + H2S —> Cu2S + 2HCl

Now we can see that the equation is balanced.

B. Determination of the number of mole of Cu2S produced by the reaction 1.2 mole of CuCl.

This is illustrated below:

2CuCl + H2S —> Cu2S + 2HCl

From the balanced equation above, 2 moles of CuCl reacted to produce 1 mole of Cu2S.

Therefore, 1.2 mole of CuCl will react to produce = 1.2/2 = 0.6 mole of Cu2S.

Therefore, 0.6 mole of Cu2S is produced from the reaction of 1.2 mole of CuCl.

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Vesna [10]

1. 12.992 L

2. 2.42 moles

3. 275.52 L

4. 567.844 g

<h3>Further explanation</h3>

Given

moles and volume at STP

Required

mass, volume and moles

Solution

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.

1. 0.58 moles ammonia :

Volume = 0.58 moles x 22.4 L = 12.992 L

2. 77.5 grams of O₂ :

Moles = 77.5 grams x (1 mol/32 grams) = 2.42

3. 12.3 mole of Bromine gas :

Volume = 12.3 mole   x (22.4 L/1 mole) = 275.52 L

4. 4.8 moles iron(II)chloride :

Mass = 4.48 moles x molar mass ( 126,751 g/mol) = 567.844 g

5 0
3 years ago
Please give me your best chemistry joke. The best joke gets Brainlest.
tia_tia [17]

Answer:

why can you never trust a atom

Explanation:

they make everything up

5 0
3 years ago
Read 2 more answers
A sample may contain any or all of the following ions: Hg22+, Ba2+, Mn2+. No precipitate formed when an aqueous solution of NaCl
Bess [88]

Answer:

Only Mn⁺² is present.

Explanation:

  • When an aqueous solution of NaCl is added, the Cl⁻ species is introduced. In the presence of Cl⁻, Hg₂²⁺ would precipitate as Hg₂Cl₂.
  • When an aqueous solution of Na₂SO₄ is added, the SO₄⁻² species is introduced. In the presence of SO₄⁻², Ba⁺² would precipitate as BaSO₄.

No precipitate formed when either of these solutions were added, thus <u>the sample does not contain Hg₂⁺² nor Ba⁺²</u>.

  • Under basic conditions Mn⁺² would precipitate as Mn(OH)₂. A precipitate formed once the solution was made basic, so <u>the sample contains Mn⁺²</u>.
3 0
3 years ago
Find the empirical formula of the compound which contains 67.6% Hg, 10.8% S, and 21.6% O.
WINSTONCH [101]

Explanation:

Divide 67.6% by 201 of mercury

Divide 10.8% by 32 of sulfur

Divide 21.6% by 16 of oxygen

Then divide number by itself .

Example : Divide 2.97 by both sulfer moles and oxygen moles . Multiply by 2 to get whole number

7 0
3 years ago
How many moles of nitrogen gas are present in a 28.9 dm3 temperature of 55.0 °C and a pressure of 144 kPa?
jeka57 [31]

Given :

Temperature, T = 55° C = ( 55 + 273 ) K = 328 K .

Volume of container, V = 28.9 dm³ = 0.0289 m³ .

Pressure, P = 144 kPa = 144000 Pa .

To Find :

Number of moles of nitrogen  gas.

Solution :

We know, by ideal gas equation :

PV = nRT  ( R ( universal gas constant ) = 8.31 J K⁻¹ mol⁻¹ .

n = \dfrac{PV}{RT}\\\\n = \dfrac{144000\times 0.0289}{8.31 \times 328} \ moles\\\\n = \dfrac{4161.6}{2725.68}\\\\n = 1.53 \ moles

Therefore, number of moles of nitrogen gas is 1.53 moles.

8 0
3 years ago
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