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3 years ago

C12H26O + SO3+NaOH ----> C12H25NaSO4+ H2O

Chemistry

1 answer:

Vlad [161]3 years ago

4 0

Answer:

We can solve this by the method of which i solved your one question earlier

so again here molar mass of C12H25NaSO4 is 288.372 and number of moles for 11900 gm of C12H25NaSO4 will be = 11900/288.372

which is almost = 41.26 moles

so to get one mole of C12H25NaSO4 we need one mole of C12H26O

so for 41.26 moles of C12H25NaSO4 it will require 41 26 moles of C12H26O

so the mass of C12H26O = 41.26× its molar mass

C12H26O = 41.26×186.34

= 7688.38 gm!!

so the conclusion is If you need 11900 g of C12H25NaSO4 (Sodium Lauryl Sulfate) you need C12H26O 7688.38 gm !!

Again i d k wether it's right or wrong but i tried my best hope it helped you!!

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CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?

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<h3>Answer:</h3>

64 g O₂

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

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<u>Step 1: Define</u>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

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[Solve] x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

[PT] Molar Mass of O - 16.00 g/mol

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

63.929 g O₂ ≈ 64 g O₂

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