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Vlad1618 [11]
3 years ago
10

Copper sulfate is a blue solid that is used to control algae growth. Solutions of copper sulfate that come in contact with the s

urface of galvanized ( Zinc-plated) steel pails undergo the following reaction that forms copper metal on the zinc surface. How many grams of Zinc would react with 454g (1lb) of copper sulfate (160g/mol)?
CuSO4(aq)+ Zn(s)>>>>Cu(s) + ZnSO4(aq)
Chemistry
1 answer:
True [87]3 years ago
4 0

Answer:

185.49 grams of Zinc would react with 454g (1lb) of copper sulfate

Explanation:

Yo know the following balanced reaction:

CuSO₄(aq)+ Zn(s) →Cu(s) + ZnSO₄(aq)

You can see that by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents and products are part of the reaction:

  • CuSO₄: 1 mole
  • Zn: 1 mole
  • Cu: 1 mole
  • ZnSO₄: 1 mole

Being:

  • Cu: 63.54 g/mole
  • S: 32 g/mole
  • O: 16 g/mole
  • Zn: 65.37 g/mole

the molar mass of the compounds participating in the reaction is:

  • CuSO₄:63.54 g/mole + 32 g/mole + 4*16 g/mole= 159.54 g/mole ≅ 160 g/mole
  • Zn: 65.37 g/mole
  • Cu: 63.54 g/mole
  • ZnSO₄: 65.37 g/mole + 32 g/mole + 4*16 g/mole= 161.37 g/mole

Then, by stoichiometry of the reaction, the following amounts of mass of reagent and product participate in the reaction:

  • CuSO₄: 1 moles* 160 g/mole= 160 g
  • Zn: 1 mole* 65.37 g/mole= 65.37 g
  • Cu: 1 mole* 63.54 g/mole= 63.54 g
  • ZnSO₄: 1 mole* 161.37 g/mole= 161.37 g

Now you can apply the following rule of three: if 160 grams of CuSO₄ react with 65.37 grams of Zn by this reaction stoichiometry, 454 grams of CuSO₄ with how much mass of Zn will it react?

mass of Zn=\frac{454 grams of CuSO_{4} *65.37 grams of Zn}{160 grams of CuSO_{4}}

mass of Zn= 185.49 grams

<u><em>185.49 grams of Zinc would react with 454g (1lb) of copper sulfate</em></u>

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Answer:

<u />

  • <u>a) 1.44g</u>

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Explanation:

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

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

<u>2. Mole ratio</u>

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

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

    Molarity=\text{number of moles of solute}/\text{volume of solution in liters}

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

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

<u>4. Find the number of grams of KHP that reacted</u>

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:

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<h3>Answer:</h3>

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

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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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  2. Parenthesis
  3. Exponents
  4. Multiplication
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<u>Step 1: Define</u>

103.4 g H₂S (Sulfuric Acid)

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

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<u>Step 3: Convert</u>

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<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

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