Answer in two or three sentences
1 answer:
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Answer:
Explanation:
From the statement of the problem,
B₂S₃ + H₂O
→ H₃BO₃
+ H₂S
B₂S₃ + H₂O → H₃BO₃ + H₂S
We that the above expression does not conform with the law of conservation of mass:
To obey the law, we need to derive a balanced reaction equation:
Let us use the mathematical method to obtain a balanced equation.
let the balanced equation be:
aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S
where a, b, c and d will make the equation balanced.
Conservating B: 2a = c
S: 3a = d
H: 2b = 3c + 2d
O: b = 3c
if a = 1,
c = 2,
b = 6,
2d = 2(6) - 3(2) = 6, d = 3
Now we can input this into our equation:
B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S
B₂S₃ + 6H₂O
→ 2H₃BO₃
+ 3H₂S
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>