Hydrogen and Oxygen by themselves have 1 and 6 valence electrons, respectively. 8 valence electrons is stable, so the atoms form bonds with each other to achieve 8 valence e-.
1 H atom + 1 H atom + 1 O atom = 8 valence e-
Answer: 6 moles of NaCl are produced when 2 moles of sodium phosphate reacts with 3 moles of calcium chloride
Explanation:
The balanced chemical equation is:
According to stoichiometry :
2 moles of 
require 3 moles of 
Thus both are limiting reagent as both will limit the formation of product.
As 2 moles of reacts with 3 moles of give = 6 moles of 
Thus 6 moles of NaCl are produced when 2 moles of sodium phosphate reacts with 3 moles of calcium chloride
1. The empirical formula of the hydrocarbon is CH₃
2. The molecular formula of the hydrocarbon is C₂H₆
<h3>How to determine the mass of Carbon </h3>
- Mass of CO₂ = 1.47 g
- Molar mass of CO₂ = 44 g/mol
- Molar of C = 12 g/mol
- Mass of C =?
Mass of C = (12 / 44) × 1.47
Mass of C = 0.4 g
<h3>How to determine the mass of H</h3>
- Mass of compound = 0.5 g
- Mass of C = 0.4 g
- Mass of H = ?
Mass of H = (mass of compound) – (mass of C)
Mass of H = 0.5 – 0.4
Mass of H =0.1 g
<h3>1. How to determine the empirical formula </h3>
- C = 0.4 g
- H = 0.1 g
- Empirical formula =?
Divide by their molar mass
C = 0.4 / 12 = 0.03
H = 0.1 / 1 = 0.1
Divide by the smallest
C = 0.03 / 0.03 = 1
H = 0.1 / 0.03 = 3
Thus, the empirical formula of the compound is CH₃
<h3>2. How to determine the molecular formula</h3>
- Empirical formula = CH₃
- Molar mass = 30 g/mol
- Molecular formula =?
Molecular formula = empirical × n = mass number
[CH₃]n = 30
[12 + (3×1)]n = 30
15n = 30
Divide both side by 15
n = 30 / 15
n = 2
Molecular formula = [CH₃]n
Molecular formula = [CH₃]₂
Molecular formula = C₂H₆
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The answer is: "FOUR (4)" .
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Answer:
88.50 %
The balance chemical equation is as follow,
2 H₂ + O₂ → 2 H₂O
Step 1: Find the limiting reactant;
According to eq.
4.032 g (2 mole) H₂ reacts with = 32 g (1 mole) of O₂
So,
11 g of H₂ will react with = X g of O₂
Solving for X,
X = (11 g × 32 g) ÷ 4.032 g
X = 87.301 g of O₂
Therefore, H₂ is the limiting reactant as O₂ is present in excess.
Step 2: Calculating %age Yield;
According to eq.
4.032 g (2 mole) H₂ produces = 36.032 g (1 mole) of H₂O
So,
11 g of H₂ will react with = X g of H₂O
Solving for X,
X = (11 g × 36.032 g) ÷ 4.032 g
X = 98.301 g of H₂O
So,
Actual Yield = 87 g
Theoretical Yield = 98.301 g
Using formula = Actual Yield / Theoretical Yield × 100
= 87 g / 98.301 × 100
= 88.50 %
