Answer: Yes, there is sufficient sodium carbonate available.
Explanation:
In this scenario, according to the specified reaction:
Using stoichiometry, one can figure out the grams of sodium carbonate required to neutralize 1,665 g of sulfuric acid as outlined below:
Hence, the amount on hand is 2.0 kg, which leaves 0.2 kg as surplus, therefore:
A. Yes, there is sufficient sodium carbonate available.
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Answer:
Explanation:
The oxidation state corresponds to the charge of each atomic ion. An increase indicates oxidation of the element while a decrease reflects reduction of the element.
2AgCl+Zn⟶2Ag+ZnCl2
Zinc undergoes oxidation, while Ag experiences reduction.
Ag⁺ changes to Ag (oxidation state decreases), thus Ag is reduced.
Zn alters to Zn⁺² (oxidation state increases), hence Zn is oxidized.
4NH₃+3O₂⟶2N₂+6H₂O
The oxidation state of nitrogen in ammonia is -3
whereas it is zero in elemental nitrogen.
An increase in the oxidation state indicates nitrogen is oxidized.
The oxidation state of oxygen is zero when in molecular oxygen and -2 when in water. Therefore, the oxidation state decreases, indicating oxidation is reduced.
Fe₂O₃+2Al⟶Al₂O₃+2Fe
The oxidation state of Fe in Fe₂O₃ is +3, switching to zero in Fe, so iron is reduced.
Aluminum's oxidation state is zero in Al, rising to +3 in Al₂O₃, indicating it is oxidized.
In a 100 g sample of the compound, there are 63.57 g of carbon, 6 g of hydrogen, 9.267 g of nitrogen, and 21.17 g of oxygen. First, convert these masses into moles (n) using the formula n = m/M, where M is the molar mass from the periodic table.
For carbon: 63.57 g C -> 63.57 g C / 12.01 g/mol = 5.29 moles C.
For hydrogen: 6 g H -> 6 g H / 1.008 g/mol = 5.95 moles H.
For nitrogen: 9.267 g N -> 9.267 g N / 14.01 g/mol = 0.6615 moles N.
For oxygen: 21.17 g O -> 21.17 g O / 16.00 g/mol = 1.32 moles O.
Thus, the mole ratio looks like this: C 5.29 H 5.95 N 0.6615 O 1.32.
Now, divide each value by the smallest number (1.32): C 4 H 4.5 N 0.5 O 1.
To eliminate fractions, multiply all values by 2, yielding C8H9N1O2.
Now, all numbers are integers! Hence, the empirical formula is C8H9NO2.
Although the empirical formula isn't always the same as the molecular formula, in this instance, it corresponds to acetaminophen.
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