Fireworks owe their colors to reactions of combustion of the metals present. When Mg and Al burn, they emit a white bright light, whereas iron emits a gold light. Besides metals, oxygen is necesary for the combustion. The decomposition reactions of barium nitrate and potassium chlorate provide this element. At the same time, barium can burn emitting a green light.
(a) Barium nitrate is a <em>salt</em> formed by the <em>cation</em> barium Ba²⁺ and the <em>anion</em> nitrate NO₃⁻. Its formula is Ba(NO₃)₂. Potassium chlorate is a <em>salt</em> formed by the <em>cation</em> potassium K⁺ and the <em>anion</em> chlorate ClO₃⁻. Its formula is KClO₃.
(b) The balanced equation for the decomposition of potassium chloride is:
2KClO₃(s) ⇄ 2KCl(s) + 3O₂(g)
(c) The balanced equation for the decomposition of barium nitrate is:
Ba(NO₃)₂(s) ⇄ BaO(s) + N₂(g) + 3O₂(g)
(d) The balanced equations of metals with oxygen to form metal oxides are:
- 2 Mg(s) + O₂(g) ⇄ 2 MgO(s)
- 4 Al(s) + 3 O₂(g) ⇄ 2 Al₂O₃(s)
- 4 Fe(s) + 3 O₂(g) ⇄ 2 Fe₂O₃(s)
Nitrogen can form a diatomic molecule held together by triple bonds.
Answer:
Percent yield = 89.1%
Explanation:
Based on the equation:
Cl₂ + 2KI → 2KCl + I₂
<em>1 mole of Cl₂ reacts with 2 moles of KI to produce to moles of KCl</em>
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To solve this quesiton we must find the moles of each reactant in order to find the limiting reactant. With the limiting reactant we can find the moles of KCl and the mass:
<em>Moles Cl₂:</em>
8x10²⁵ molecules * (1mol / 6.022x10²³ molecules) = 133 moles
<em>Moles KI -Molar mass: 166.0028g/mol-</em>
25g * (1mol / 166.0028g) = 0.15 moles
Here, clarely, the KI is the limiting reactant
As 2 moles of KI produce 2 moles of KCl, the moles of KCl produced are 0.15 moles. The theoretical mass is:
0.15 moles * (74.5513g / mol) =
11.2g KCl
Percent yield is: Actual yield (10.0g) / Theoretical yield (11.2g) * 100
<h3>Percent yield = 89.1%</h3>
Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
