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Citrus2011 [14]
2 years ago
7

80g of sodium hydroxide reacts with excess sulfuric acid to produce 142g of sodium sulfate. What mass of sodium hydroxide would

be needed in order to make 50g of sodium sulfate?
Chemistry
1 answer:
Ostrovityanka [42]2 years ago
6 0

Answer:

Mass of sodium hydroxide needed  = 28.2 g

Explanation:

Given data:

Mass of sodium hydroxide = 80 g

Mass of sodium sulfate produced = 142 g

Mass of sodium hydroxide needed = ?

Mass of sodium sulfate produced = 50 g

Solution:

Mass of sodium sulfate produced / Mass of sodium hydroxide = Mass of sodium sulfate produced / x

142 g/ 80 g = 50 g / x

x = 50 g × 80 g / 142 g

x =4000 g/ 142

x = 28.2 g

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3 years ago
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Lyrx [107]

Answer:

5.702 mol K₂SO₄

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Compounds
  • Moles

<u>Stoichiometry</u>

  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[Given] 993.6 g K₂SO₄

[Solve] moles K₂SO₄

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

[PT] Molar Mass of K: 39.10 g/mol

[PT] Molar Mass of S: 32.07 g/mol

[PT] Molar mass of O: 16.00 g/mol

Molar Mass of K₂SO₄: 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                       \displaystyle 993.6 \ g \ K_2SO_4(\frac{1 \ mol \ K_2SO_4}{174.27 \ g \ K_2SO_4})
  2. [DA] Divide [Cancel out units]:                                                                         \displaystyle 5.7015 \ mol \ K_2SO_4

<u>Step 4: Check</u>

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

5.7015 mol K₂SO₄ ≈ 5.702 mol K₂SO₄

7 0
3 years ago
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled
kirza4 [7]

Answer:

Nitrogen, N_2\\

Explanation:

Hello,

This is a clear example of what the ideal gas equation is used for, thus, from its mathematical definition:

PV=nRT

One can spell it out in terms of mass and molar mass:

PV=\frac{m}{M}RT

Now, solving for the molecular mass, M:

M=\frac{mRT}{PV} =\frac{9.5g*0.082\frac{atm*L}{mol*K}*297.15K}{2.00atm*4.1L}\\ M=28.23g/mol

Now, by taking into account that the gas is diatomic, the matching gas turns out to be nitrogen.

Best regards.

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2 years ago
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