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Yuri [45]
3 years ago
12

A solution is created by dissolving 13.5 grams of ammonium chloride in enough water to make 235 ml of solution. how many moles o

f ammonium chloride are present in the resulting solution?
Chemistry
1 answer:
erica [24]3 years ago
7 0
<span>Moles = 0.252 Molarity = 1.07 This question is badly worded. You're asking for moles and I suspect you really want molarity. The number of moles of ammonium chloride you have in the solution will remain constant regardless of the volume of the solution. However, the molarity of the solution will differ depending upon how concentrated it is. So I'll give you both the number of moles of ammonium chloride you have, and the molarity of the resulting solution. Please talk to your teacher if you're confused by the difference between moles and molarity. The formula for ammonium chloride is NH4Cl. So let's calculate it's molar mass. Start by looking up the associated atomic weights. Atomic weight nitrogen = 14.0067 Atomic weight hydrogen = 1.00794 Atomic weight chlorine = 35.453 Molar mass NH4Cl = 14.0067 + 4 * 1.00794 + 35.453 = 53.49146 g/mol Moles NH4Cl = 13.5 g / 53.49146 g/mol = 0.252376735 mol Molarity is defined as moles per liter, so let's divide the number of moles we have by the volume in liters. So: 0.252376735 mol / 0.235 l = 1.073943551 M Rounding to 3 significant figures gives: 0.252 moles, 1.07 molarity.</span>
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one reaction that produces hydrogen gas can be represented by the unbalanced chemical equation Mg(s)+HCI(aq) -&gt; MgCI(aq)+H2(g
Sonbull [250]
<h3>Answer:</h3>

128 g HCl

<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<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Reaction Mole Ratios
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

↓

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

[Given] 3.25 mol Mg

[Solve] x g HCl

<u>Step 2: Identify Conversions</u>

[RxN] 2 mol Mg → 2 mol HCl

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

<u>Step 3: Stoich</u>

  1. [S - DA] Set up:                                                                                                 \displaystyle 3.25 \ mol \ Mg(\frac{2 \ mol \ HCl}{2 \ mol \ Mg})(\frac{36.46 \ g \ HCl}{1 \ mol \ HCl})
  2. [S - DA] Multiply/Divide [Cancel out units]:                                                    \displaystyle 127.61 \ g \ HCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

127.61 g HCl ≈ 128 g HCl

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