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Answer:
There will be 1.50 gram of water
1. ₉¹⁹F
2. Mg(OH)₂
3. 2H₂ + O₂ ⇒ 2H₂O
<h3>Further explanation</h3>
1. Fluorine, atomic number : 9 , mass number = 19
Symbol : ₉¹⁹F
protons=electrons=atomic number = 9
neutrons = mass number - atomic mass
Configuration : [He] 2s² 2p⁵
2. Magnesium hydroxide is an ionic compound and is a strong base consisting of 2 ions:
Positive ion: Magnesium: Mg²⁺
negative ion: Hydroxide: OH⁻
The charges of the two are crossed, so that the compound becomes:
Mg(OH)₂
3. Reaction :
H₂ + O₂ --------> H₂O
give coefficient :
aH₂ + bO₂ --------> H₂O
H, left = 2a, right 2⇒2a=2⇒a=1
O, left = 2b, right 1⇒2b=1⇒b=0.5
Reaction becomes :
H₂ + 0.5O₂ --------> H₂O x 2
2H₂ + O₂ --------> 2H₂O
The percent yield of the reaction between ammonia gas with oxygen gas is 90.52%.
A chemical reaction between ammonia gas (NH3) with oxygen gas (O2)
NH₃ + O₂ → NO₂ + H₂O
The balanced reaction 4NH₃ + 7O₂ → 4NO₂ + 6H₂O
Calculate the number of moles from the reactant
- Ammonia gas
Molar mass N = 14 gr/mol
Molar mass H = 1 gr/mol
Molar mass NH₃ = 14 + (3 × 1) = 14 + 3 = 17 gr/mol
mass = 28.5 grams
n = m ÷ molar mass = 28.5 ÷ 17 = 1.68 mol - Oxygen gas
Molar mass O = 16 gr/mol
Molar mass O₂ = 16 × 2 = 32 gr/mol
mass = 83.4 grams
n = m ÷ molar mass = 83.4 ÷ 32 = 2.61 mol - n O₂ ÷ coefficient O₂ = 2.61 ÷ 7 = 0.37
n NH₃ ÷ coefficient NH₃ = 1.68 ÷ 4 = 0.42
0.42 > 0.37 it means that the ammonia gas is in excess and the O₂ is limiting.
According to stoichiometry, the number of moles NO₂ with the number of moles O₂ has the ratio with the coefficient in reaction.
- Theoretically the number moles of NO₂
n O₂ : n NO₂ = 7 : 4
2.61 : n NO₂ = 7 : 4
n NO₂ = 4 x 2.61 : 7 = 1.49 mol - The actual number of moles NO₂
Molar mas NO₂ = 14 + (16 × 2) = 14 + 32 = 46 gr/mol
n NO₂ = m ÷ molar mass = 61.9 ÷ 46 = 1.35 mol
The percent yield NO₂ is the ratio of the actual number of moles NO₂ with the theoretical number of moles NO₂ times 100%.
P = (1.35 ÷ 1.49) × 100%
P = 0.9052 × 100%
P = 90.52%
Learn more about stoichiometry here: brainly.com/question/13691565
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