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dlinn [17]
1 year ago
5

g how many grams of chlorine gas are produced in the following reaction from 0.490 g hydrogen and 50.0 g chlorine? which substan

ce is the limiting reactan
Chemistry
1 answer:
Fittoniya [83]1 year ago
7 0

The formed HCl concentration is 17.6g, and the limiting reactant is hydrogen.

<h3>What is limiting reactant?</h3>

In this case, based on the information provided, we can calculate the required grams of HCl by first identifying the limiting reactant using the moles of each reactant, which are in a 1:1 mole ratio:

nH₂ = .490gH₂ x 1 mol H₂ / 2.02g H₂ = 0.242 mol H₂

nCl₂ = 50.0gCl₂ x 1 mol Cl₂ / 70.9g Cl₂ = 0.705mol Cl₂

As a result, we conclude that hydrogen is the limiting reactant, and we use a 1:2 mole ratio with HCl, whose molar mass is 36.46 g/mol:

mHCl = .242 mol H₂ x 2mol HCl / 1 mol H₂ x 36.6g HCl / 1mol HCl

mHCl = 17.6g HCL

Therefore the concentration of HCl formed is 17.6g  and the limiting reactant is hydrogen.

To learn more about reactions refer to :

brainly.com/question/24795637

#SPJ4

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If a gas occupies 1532.7 mL at standard temperature, what volume does it occupy at 49.4 ºC if the pressure remains constant?
ElenaW [278]

Answer:

a. 1810mL

Explanation:

When conditions for a gas change under constant pressure (and the number of molecules doesn't change), it follows Charles' Law:

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}  where the temperatures must be measured in Kelvin

To convert from Celsius to Kelvin, add 273, or use the equation:  T_C+273=T_K

For this problem, one must also recall that standard temperature is 0°C (or 273K).

So, T_1 = 273[K], and T_2 = (49.4+273)[K]=322.4[K].

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

\dfrac{(1532.7[mL])}{(273[K])}=\dfrac{V_2}{(322.4[K])}

\dfrac{(1532.7[mL])}{(273[K\!\!\!\!\!{-}])}(322.4[K\!\!\!\!\!{-}] )=\dfrac{V_2}{(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})}(322.4[K]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{----})

1810.04571428[mL]=V_2

Adjusting for significant figures, this gives V_2=1810[mL]

4 0
1 year ago
A gas with a volume of 4.0 L at a pressure of 2.02 atm is allowed to expand to a volume of 12.0
Maurinko [17]
<h3>Answer:</h3>

P₂ = 0.67 atm

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Gas Laws</u>

Boyle's Law: P₁V₁ = P₂V₂

  • P₁ is pressure 1
  • V₁ is volume 1
  • P₂ is pressure 2
  • V₂ is volume 2
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] P₁ = 2.02 atm

[Given] V₁ = 4.0 L

[Given] V₂ = 12.0 L

[Solve] P₂

<u>Step 2: Solve</u>

  1. Substitute in variables [Boyle's Law]:                                                              (2.02 atm)(4.0 L) = P₂(12.0 L)
  2. [Pressure] Multiply:                                                                                           8.08 atm · L = P₂(12.0 L)
  3. [Pressure] [Division Property of Equality] Isolate unknown:                          0.673333 atm = P₂
  4. [Pressure] Rewrite:                                                                                           P₂ = 0.673333 atm

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs as our smallest.</em>

0.673333 atm ≈ 0.67 atm

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Explanation:

The states may differ depending on the reactions

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harkovskaia [24]

Answer:

percentage by mass of each element in a compound.

Explanation:

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2 years ago
Could somebody please give me an example of some physical properties for any substance?
KIM [24]

Answer:

A physical property is a characteristic of matter that is not associated with a change in its chemical composition. Familiar examples of physical properties include density, color, hardness, melting and boiling points, and electrical conductivity.

Explanation:

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