Answer:
144g of H₂O
Explanation:
3NH₄ClO₄(s) + 3Al → Al₂O₃(s) + AlCl₃(s) + 3NO(g) + 6H₂O(g)
From the equation:
3 moles of NH₄ClO₄ produced 6 moles of H₂O
4 moles of NH₄ClO₄ produced ? moles of H₂O
(4 ₓ 6)/3 =
= 8 moles of H₂O
1 mole of H₂O = (1 × 2) + 16 = 18g (The Relative Molecular mass of H₂O)
8 moles of H₂O = ?
Therefore 8 × 18 = 144g
=144g of H₂O
The reactions are in order which includes combustion reaction, Hydration reaction, oxidation reaction, and displacement reaction.
a) A combustion reaction is a chemical reaction between a fuel and an oxidant where heat is released. The combustion reaction example is given below. It is a balanced chemical reaction.
2C₃H₆(g) + 9O₂(g) --------> 6CO₂(g) + 6H₂O(g)
b. A hydration reaction is a chemical reaction in which a molecule of water is added to another molecule. Here Aluminum oxide is added to water to form aluminum hydroxide.
4Al₂O3(s) + 6H₂O(l)------> 2Al(OH)3(s)
c. When a metal reacts with oxygen, the metal forms an oxide. Oxide is a compound of metal and oxygen. Here lithium metal reacts with oxygen to form lithium oxide.
2Li(s) + O₂(g)-----> Li₂O(s)
d. A displacement reaction is one in which a more reactive element displaces a less reactive element from a compound. Here Zinc is more reactive than silver, so silver was displaced to form Zinc Nitrate.
Zn(s) + 2AgNO₃(aq) -----> 2Ag(s) + Zn(NO₃)₂(aq)
To know more about reactions, click below:
brainly.com/question/11231920
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Answer:
The answer to your question is
1.-Fe₂O₃
2.- 280 g
3.- 330 g
Explanation:
Data
mass of CO = 224 g
mass of Fe₂O₃ = 400 g
mass of Fe = ?
mass of CO₂
Balanced chemical reaction
Fe₂O₃ + 3CO ⇒ 2Fe + 3CO₂
1.- Calculate the molar mass of Fe₂O₃ and CO
Fe₂O₃ = (56 x 2) + (16 x 3) = 160 g
CO = 12 + 16 = 28 g
2.- Calculate the proportions
theoretical proportion Fe₂O₃ /3CO = 160/84 = 1.90
experimental proportion Fe₂O₃ / CO = 400/224 = 1.78
As the experimental proportion is lower than the theoretical, we conclude that the Fe₂O₃ is the limiting reactant.
3.- 160 g of Fe₂O₃ --------------- 2(56) g of Fe
400 g of Fe₂O₃ --------------- x
x = (400 x 112) / 160
x = 280 g of Fe
4.- 160 g of Fe₂O₃ --------------- 3(44) g of CO₂
400 g of Fe₂O₃ -------------- x
x = (400 x 132)/160
x = 330 gr