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:
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>
C. CaCl2 is the correct answer
About 6.5x10^22 molecules.
(5g C2H5OH)x(1 mol C2H5OH/46g C2H5OH)x(6.02x10^23 molecules C2H5OH/1 mol C2H5OH)=(3.01E24)/46=6.5x10^22.
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Answer:
Explanation:
Here we have to use stoichiometry.
First of all, we have to calculate the mass of 100% of yield:
1.7 g ------- 98%
X -------- 100%
X = 1.73 g (approximately)
Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.
Molar Mass N2 : 14x2 = 28 g/mol
Molar Mass NH3: 14 + 3 = 17 g/mol
28g (N2) ------- 17x2 (NH3)
X ------------ 1.73 g
X = 1.42 g (approximately)