<span>When the hexene is treated with an excess of hydrogen gas and a platinum catalyst, the product formed is
hexane.
Hexene is an unsaturated hydrocarbon. Hexene when treated with H2 gas, in presence with Pt catalyst, the double bond is broken. This results in generation of saturated hydrocarbon i.e. hexane. </span>
<u>Answer:</u> The gas produced when sodium phosphide reacts with water is phosphine.
<u>Explanation:</u>
When sodium phosphide reacts with water molecule, it leads to the production of flammable, poisonous gas known as phosphine along with the production of sodium hydroxide.
The chemical reaction for the reaction of sodium phosphide with water follows the equation:
By Stoichiometry of the reaction:
1 mole of sodium phosphide reacts with 3 moles of water to produce 1 mole of phosphine gas and 3 moles of sodium hydroxide.
Hence, the gas produced when sodium phosphide reacts with water is phosphine.
Answer:
iron (III) oxide =Fe2O3
iron(III) carbonate =Fe2(CO3)3
The mass in grams of NH₃ produced from the reaction is 3.4 g
<h3>Balanced equation</h3>
We'll begin by writing the balanced equation for the reaction. This illustrated below:
N₂ + 3H₂ -> 2NH₃
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
<h3>How to determine the volume of NH₃ produced</h3>
From the balanced equation above,
1 dm³ of N₂ reacted to produced 2 dm³ NH₃
Therefore,
2.24 dm³ of N₂ will react to produce = 2.24 × 2 = 4.48 dm³ of NH₃
<h3>How to determine the mass of NH₃ produced</h3>
We'll begin by obtained the mole of 4.48 dm³ of NH₃. Details below:
22.4 dm³ = 1 mole NH₃
Therefore,
4.48 dm³ = 4.48 / 22.4
4.48 dm³ = 0.2 mole of NH₃
Finally, we shall determine the mass of NH₃ as follow:
- Molar mass of NH₃ = 17 g/mol
- Mole of NH₃ = 0.2 mole
- Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 0.2 × 17
Mass of NH₃ = 3.4 g
Learn more about stoichiometry:
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The compound that has the smallest ions with the greatest charge will have the strongest ionic bonds. On the other hand, large and minimally charged ions will form weaker bonds.
