As fuel is burned in an airplane, it seems to disappear from the tank, suggesting that matter is destroyed during the reaction.
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The number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
<h3>What are moles?</h3>A mole is defined as 6.02214076 × of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.
Given data:
Moles of hydrochloric acid = 0.385 mol
Mass of chlorine gas =?
Chemical equation:
4HCl + O₂ → 2Cl₂ + 2H₂O
Now we will compare the moles of Cl₂ with HCl.
HCl : Cl₂
4 : 2
0.385 : 2÷4× 0.385 = 0.1925 mol
Oxygen is present in excess that's why the mass of chlorine produced depends upon the available amount of HCl.
Mass of Cl₂ :
Mass of Cl₂ = moles × molar mass
Mass of Cl₂ =0.1925 mol × 71 g/mol
Mass of Cl₂ = 13.6675 g
Hence, the number of grams of Cl2 formed when 0.385 mol HCl reacts with an excess of O2 is 13.6675 g.
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Answer:
The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.
Explanation:
You know the balanced reaction:
4 NA + O₂ ⟶ 2 Na₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:
- Na: 4 moles
- O₂: 1 mole
- Na₂O: 2 moles
Being:
- Na: 23 g/ole
- O: 16 g/mole
the molar mass of the compounds participating in the reaction is:
- Na: 23 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- Na₂O: 2*23 g/mole +16 g/mole= 62 g/mole
Then by stoichiometry of the reaction they react and are produced:
- Na: 4 moles* 23 g/mole= 92 g
- O₂: 1 mole*32 g/mole= 32 g
- Na₂O: 2 moles* 62 g/mole= 124 g
Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?
mass of Na₂O=5.39 g
<em><u>The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.</u></em>