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tekilochka [14]
3 years ago
8

using the equation 2h2+o2-->2h2o if 192g of oxygen are produced how many grmas of hydrogen must react with it

Chemistry
2 answers:
OleMash [197]3 years ago
8 0
Mol H2O = weight H2O / MW H2O = 1.8 / 18 = 0.1 mol
Weight H2 = mol H2 x MW H2
1 Mol H2 --> 1 mol H2O
So :
Weight H2 = mol H2O x MW H2 = 0.1 x 2 = 0.2 grams
Brilliant_brown [7]3 years ago
7 0
Equation: 2H₂ + O₂ → 2H₂O

Now, Given mass of Oxygen = 192 g
Molar mass of Oxygen = 16 g/mol

No. of moles in Oxygen = 16/192 = 0.0833

Now, for every mole of Oxygen, 2 mole of Hydrogen will form, 
so, Number of moles of Hydrogen = 0.0833 * 2 = 0.167

Given mass = Number of Moles * Molar mass
Given mass = 0.167 * 2
m = 0.33 g

In short, Your Answer would be: 0.33 g

Hope this helps!
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2 NaOH (s) + CO2(g) → Na2CO3 (s) + H20 (I)
Paha777 [63]
<h3>Answer:</h3>

16.7 g H₂O

<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
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<u>Chemistry</u>

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] 2NaOH (s) + CO₂ (g) → Na₂CO₃ (s) + H₂O (l)

[Given] 1.85 mol NaOH

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

[RxN] 2 mol NaOH → 1 mol H₂O

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Molar Mass of O - 16.00 g/mol

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<u>Step 3: Stoichiometry</u>

  1. Set up:                               \displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})
  2. Multiply/Divide:                 \displaystyle 16.6685 \ g \ H_2O

<u>Step 4: Check</u>

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

16.6685 g H₂O ≈ 16.7 g H₂O

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