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Mice21 [21]
2 years ago
6

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

situation: You combine 0.5 grams of Na2CO3 with excess CaCl2. How many grams of NaCl would you expect this reaction to produce? Show all work below.
Chemistry
1 answer:
LenKa [72]2 years ago
5 0
<h3>Answer:</h3>

0.6 g NaCl

<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>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

  1. Set up:                    \displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})
  2. Multiply/Divide:                                                                                               \displaystyle 0.551373 \ g \ NaCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

0.551373 g NaCl ≈ 0.6 g NaCl

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Answer:

<u />

  • <u>a) 1.44g</u>

<u />

  • <u>b) 77.3%</u>

<u />

Explanation:

<u>1. Chemical balanced equation (given)</u>

       NaOH(aq)+ KHP(s)\rightarrow NaKP(aq)+H_2O(l)

<u>2. Mole ratio</u>

1molNaOH(aq):1molKHP(s)

This is, 1 mol of NaOH will reacts with 1 mol of KHP.

<u />

<u>3. Find the number of moles in 72.14 mL of the base</u>

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    \text{Volume of solution}=72.14mL=0.07214liters

     \text{Number of moles of NaOH}=0.0978M\times 0.07214liter=0.007055mol

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The number of moles of KHP that reacted is equal to the number of moles of NaOH, 0.007055 mol

Convert moles to grams:

  • mass = number moles × molar mass = 0.007055mol × 204.23g/mol
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You have to round to 3 significant figures: 1.44 g (because the molarity is given with 3 significant figures).

<u>5. Find the percentage of KHP in the sample</u>

The percentage is how much of the substance is in 100 parts of the sample.

The formula is:

  • % = (mass of substance / mass of sample) × 100

  • % = (1.4408g/ 1.864g) × 100 = 77.3%
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