<u>Answer:</u> The enthalpy of the reaction for given amount of aluminium will be 
<u>Explanation:</u>
We are given:
Moles of aluminium = 12 moles
For the given chemical reaction:
By Stoichiometry of the reaction:
If 2 moles of aluminium produces -850 kJ of energy.
Then, 12 moles of aluminium will produce = 
of energy.
Converting this into three significant figures, we get:
Thus, the enthalpy of the reaction for given amount of aluminium will be
Answer:
The answer to your question is the letter d.
Explanation:
Data
Zn + HCl ⇒ ZnCl₂ + H₂
- Calculate the Oxidation number of reactants and products
Zn⁰ + H⁺¹Cl⁻¹ ⇒ Zn⁺²Cl₂⁻¹ + H₂⁰
- Determine which elements changed their oxidation numbers
Zn⁰ ⇒ Zn⁺² Zn lost 2e⁻
2H⁺¹ ⇒ H₂⁰ H won 2 e⁻
-Interchange the electrons lost and won
2( Zn⁰ ⇒ Zn⁺²)
2(2H⁺¹ ⇒ H₂⁰)
-Simplify
2Zn⁰ ⇒ 2Zn⁺²
4H⁺¹ ⇒ 2H₂⁰
-Substitute the coefficient in the equation
2Zn + 4HCl ⇒ 2ZnCl₂ + 2H₂
Reactant Element Product
2 Zn 2
4 H 4
4 Cl 4
- Now the reaction is balanced
Answer:
2 KCIO₃ → 2 KCI + 3 O₂
Explanation:
A chemical equation is balanced when the number of atoms of each element is equal on both sides of the equation.
You have already identified the initial number of atoms of each element on both sides of the equation. As a rule of thumb, we balance the number of oxygen and hydrogen atoms last.
However, since all the other elements are already balanced, we shall start by balancing the number of oxygen atoms.
The lowest common multiple of 2 and 3 is 6. Thus, we shall ensure that both sides of the equation has 6 oxygen atoms.
2 KCIO₃ → KCI + 3 O₂
<u>Reactants</u>
K --- 2
C--- 2
I --- 2
O --- 6
<u>Products</u>
K --- 1
C --- 1
I --- 1
O --- 6
Notice that number of K, C and I atoms on the left-hand side of the equation has also changed.
2 KCIO₃ → 2 KCI + 3 O₂
<u>Reactants</u>
K --- 2
C --- 2
I --- 2
O --- 6
<u>Products</u>
K --- 2
C --- 2
I --- 2
O --- 6
The equation is now balanced.
34.5 X 10^-11 grams of lead
