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vaieri [72.5K]
2 years ago
12

How many grams of fe2o3 are present in 0.82 mol?

Chemistry
2 answers:
muminat2 years ago
6 0

Answer:- There are 131 g of Fe_2O_3 in 0.82 moles of it.

Solution:- It is a mol to gram conversion. We have been given with 0.82 moles of Fe_2O_3 and asked to convert it to grams.

For mol to gram conversion we multiply the moles by the molar mass.

Atomic mass of Fe is 55.85 and atomic mass of O is 16. So, molar mass of Fe_2O_3 = 2(55.85) + 3(16)

= 111.7 + 48

= 159.7(\frac{g}{mol})

Let's convert the moles to grams as:

0.82mol(\frac{159.7g}{mol})

= 131 g

So, there are 131 g of Fe_2O_3 present in 0.82 mol of it.


mars1129 [50]2 years ago
5 0
130.954g of Fe2O3 in 0.82 mol
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A student has a mixture of salt (NaCl) and sugar (C12H22O11). To determine the percent composition, the student measures out 5.8
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Answer:

<u>1. Net ionic equation:</u>

  • Cl⁻(aq) +  Ag⁺(aq) → AgCl(s)

<u />

<u>2. Volume of 1.0M AgNO₃</u>

  • 41ml

Explanation:

1. Net ionic equation for the reaction of NaCl with AgNO₃.

i) Molecular equation:

It is important to show the phases:

  • (aq) for ions in aqueous solution
  • (s) for solid compounds or elements
  • (g) for gaseous compounds or elements

  • NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

ii) Dissociation reactions:

Determine the ions formed:

  • NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
  • AgNO₃(aq) → Ag⁺(aq) + NO₃⁻(aq)
  • NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

iii) Total ionic equation:

Substitute the aqueous compounds with the ions determined above:

  • Na⁺(aq) + Cl⁻(aq) +  Ag⁺(aq) + NO₃⁻(aq) → AgCl(s) +  Na⁺(aq) + NO₃⁻(aq)

iv) Net ionic equation

Remove the spectator ions:

  • Cl⁻(aq) +  Ag⁺(aq) → AgCl(s) ← answer

2.  How many mL of 1.0 M AgNO₃ will be required to precipitate 5.84 g of AgCl

i) Determine the number of moles of AgNO₃

The reaction is 1 to 1: 1 mole of AgNO₃ produces 1 mol of AgCl

The number of moles of AgCl is determined using the molar mass:

  • number of moles = mass in grams / molar mass
  • molar mass of AgCl = 143.32g/mol
  • number of moles = 5.84g / (143.32g/mol) = 0.040748 mol

ii) Determine the volume of AgNO₃

  • molarity = number of moles of solute / volume of solution in liters

  • 1.0M = 0.040748mol / V

  • V = 0.040748mol / (1.0M) = 0.040748 liter

  • V = 0.040748liter × 1,000ml / liter = 40.748 ml

Round to two significant figures: 41ml ← answer

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