What is the mass percentage of silicon?
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The coefficients of the hydrocarbon and water is 61 and it is explained below.
Explanation:
We have to balance the chemical equation that describes the combustion of octane C₈H₁₈. The combustion of octane involves burning this hydrocarbon in the presence of excess oxygen,O 2. Because octane is a hydrocarbon, that is the compound that contains only carbon and hydrogen, the reaction will produce two products carbon dioxide CO 2 and water H 2 O.
The unbalanced chemical equation is
C₈H₁₈ + O₂ → CO₂ + H₂O
To balance this equation, we have 8 present on the reactant side, so multiply the carbon dioxide by 8 to get 8 atoms of carbon on the products side.
C₈H₁₈ + O₂ → 8 CO₂ + H₂O
Now, we have 18 on the reactant side and 2 on the products side, so multiply the water molecule by 9 to get
C₈H₁₈ + O₂ → 8 CO₂ + 9 H₂O
8 * 2 atoms O + 9 *1 atoms O = 25 atoms O
By adding fractional coefficient to O₂ we get
C₈H₁₈ + O₂ → 8 CO₂ + 9 H₂O
Multiply all the coefficients by 2 we get,
2C₈H₁₈ + ( * 2) O₂ → 8 * 2 CO₂ + 9* 2 H₂O
The balance equation is
2C₈H₁₈ + 25 O₂ → 16 CO₂ + 18 H₂O
Add the coefficients to get their sum
= 2 +25 + 16 +18 = 61
Thus, The coefficients of the hydrocarbon and water is 61
<h3>Answer:</h3>
The mass of excessive water (H₂O) is 40.815 kg
<h3>Explanation:</h3>The Equation for the reaction is;
Li₂O(s) + H₂O(l) → 2LiOH(s)
From the question;
Mass of water removed is 80.0 kg
Mass of available Li₂O is 65.0 kg
We are required to calculate the mass of excessive reagent.
<h3>Step 1: Calculating the number of moles of water to be removed</h3>Moles = Mass ÷ Molar mass
Molar mass of water = 18.02 g/mol
Mass of water = 80 kg (but 1000 g = 1kg)
= 80,000 g
Therefore;
= 4.44 × 10³ moles
<h3>Step 2: Moles of Li₂O available </h3>Moles = mass ÷ molar mass
Mass of Li₂O available = 65.0 kg or 65,000 g
Molar mass Li₂O = 29.88 g/mol
Moles of Li₂O = 65,000 g ÷ 29.88 g/mol
= 2.175 × 10³ moles Li₂O
<h3>Step 3: Mass of excess reagent </h3>From the equation; Li₂O(s) + H₂O(l) → 2LiOH(s)
1 mole of Li₂O reacts with 1 mole of water to form two moles of LiOH
The ratio of Li₂O to H₂O is 1:1
- Thus, 2.175 × 10³ moles of Li₂O will react with 2.175 × 10³ moles of water.
- However, the number of moles of water to be removed is 4.44 × 10³ moles but only 2.175 × 10³ moles will react with the available Li₂O.
- This means, Li₂O is the limiting reactant while water is the excessive reagent.
Therefore:
Moles of excessive water = 4.44 × 10³ moles - 2.175 × 10³ moles
= 2.265 × 10³ moles
Mass of excessive water = 2.265 × 10³ moles × 18.02 g/mol
= 4.0815 × 10⁴ g or
= 40.815 kg
Thus, the mass of excessive water is 40.815 kg