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Paladinen [302]
2 years ago
10

If 8cm^3 of H2 reacts with an excess of Cl2, calculate how much of the HCL(g) is produced.

Chemistry
1 answer:
Rom4ik [11]2 years ago
8 0
First, we need the balanced equation: H₂ + Cl₂ ---> 2HCl

since not much information is given, I am assuming we are at STP and that 22.4 Liters= 1 mol

1) let's convert the volume to moles using the molar volume of a gas. also we need to convert the cm₃ to mL, then to Liters.

8 cm³ (1 ml/ 1 cm³)(1 L/ 1000 mL) (1 mol/ 22.4 Liters)= 3.6x10⁻⁴ moles of H₂

2) let's use the mole ratio of the balanced equation to convert moles of H₂ to moles of HCl

3.6x10⁻⁴ mol H₂ (2 mol HCl/ 1 mol H₂)= 7.1x10⁻⁴ mol HCl

3) lastly, we convert the moles of HCl to grams using the molar mass.

molar mass of HCl= 1.01 + 35.5= 36.51 g/mol

7.1x10⁻⁴ mol HCl (36.51 g/mol)=<span> 0.026 grams HCl</span>
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3Cu + 8HNO = 3Cu(NO3)2 +2NO + 4H2O
AysviL [449]

Taking into account the reaction stoichiometry, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

3 Cu+ 8 HNO₃ → 3 Cu(NO₃)₂ + 2 NO + 4 H₂O

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • Cu: 3 moles
  • HNO₃: 8 moles
  • Cu(NO₃)₂: 3 mole
  • NO: 2 moles
  • H₂O: 4 moles

The molar mass of the compounds is:

  • Cu: 63.55 g/mole
  • HNO₃: 63 g/mole
  • Cu(NO₃)₂: 187.55 g/mole
  • NO: 30 g/mole
  • H₂O: 18 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • Cu: 3 moles ×63.55 g/mole= 190.65 grams
  • HNO₃: 8 moles ×63 g/mole= 504 grams
  • Cu(NO₃)₂: 3 moles ×187.55 g/mole= 562.65 grams
  • NO: 2 moles ×30 g/mole= 60 grams
  • H₂O: 4 moles ×18 g/mole= 72 grams

<h3>Mass of Cu(NO₃)₂ produced</h3>

The following rule of three can be applied: if by reaction stoichiometry 504 grams of HNO₃ form 562.65 grams of Cu(NO₃)₂, 4.69 grams of HNO₃ form how much mass of Cu(NO₃)₂?

mass of Cu(NO_{3} )_{2} =\frac{4.69 grams of HNO_{3}  x562.65 grams of Cu(NO_{3} )_{2} }{504 grams of HNO_{3} }

<u><em>mass of Cu(NO₃)₂=  5.2634 grams</em></u>

Then, 5.2634 grams of Cu(NO₃)₂ are formed when 4.69 grams of HNO₃, assuming an excess of solid copper is present.

Learn more about the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

#SPJ1

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gregori [183]

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E. Wrong because the equation did not start with a single compound and break down. ex. (xy > x + y)

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