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3 years ago

How many moles of carbon dioxide gas, co2, are there in 2.94 l of this gas at stp

Chemistry

2 answers:

Colt1911 [192]3 years ago

7 0

Pls go through this solution

OlgaM077 [116]3 years ago

7 0

Please see workings in photo! 0.131mol CO2 at STP

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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

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Stoichiometry</u>

Reading a Periodic Table
Using Dimensional Analysis

<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

Molar Mass of H - 1.01 g/mol

Molar Mass of O - 16.00 g/mol

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

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 1.85 \ mol \ NaOH(\frac{1 \ mol \ H_2O}{2 \ mol \ NaOH})(\frac{18.02 \ g \ H_2O}{1 \ mol \ H_2O})$
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

6 0

3 years ago

N2(g) + 3H2(g) → 2NH3(g) How many grams of N2 are required to produce 240.0g NH3?

just olya [345]

Answer:

$\large \boxed{\text{197.4 g}}$

Explanation:

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ: 28.01 17.03

N₂ + 3H₂ ⟶ 2NH₃

m/g: 240.0

(a) Moles of NH₃

$\text{Moles of NH}_{3} = \text{240.0 g NH}_{3}\times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}}= \text{14.09 mol NH}_{3}$

(b) Moles of N₂

$\text{Moles of N$_{2}$} = \text{14.09 mol NH}_{3} \times \dfrac{\text{1 mol N$_{2}$}}{\text{2 mol NH}_{3}} = \text{7.046 mol N$_{2}$}$

(c) Mass of N₂

$\text{Mass of N$_{2}$} =\text{7.046 mol N$_{2}$} \times \dfrac{\text{28.01 g N$_{2}$}}{\text{1 mol N$_{2}$}} = \textbf{197.4 g N$_{2}$}\\\\\text{The reaction requires $\large \boxed{\textbf{197.4 g}}$ of N$_{2}$}$

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Answer: beta, gamma, alpha

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