If 4.80 mol Ca mixed with 2 mol N2, which is the limiting reactant? 3Ca (s) + N2 (g) Ca3N2 (s)
1 answer:
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Answer:
empirical formula:
2 g H
32.7 g S
65.3 g O
Explanation:
Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:
- 2.00% * 100g = 2 g H
- 32.7% * 100g = 32.7 g S
- 65.3% * 100g = 65.3 g O
Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:
- the molar mass of H is 1.01 g/mol
- the molar mass of S is 32.06 g/mol
- the molar mass of O is 16 g/mol
2 g H ÷ 1.01 g/mol = 1.98 mol H
32.7 g S ÷ 32.06 g/mol = 1.02 mol S
65.3 g O ÷ 16 g/mol = 4.08 mol O
Our ratio of H : S : O is now:
1.98 mol : 1.02 mol : 4.08 mol
Divide them all by the smallest number, which is 1.02:
1.98/1.02 : 1.02/1.02 : 4.08/1.02
1.94 : 1 : 4
1.94 ≈ 2
So:
2 : 1 : 4
Thus, the empirical formula is: .
<u>Answer:</u> The number of carbon and oxygen atoms in the given amount of carbon dioxide is and
respectively
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Given mass of carbon dioxide gas = 3.45 g
Molar mass of carbon dioxide gas = 44 g/mol
Putting values in above equation, we get:
1 mole of carbon dioxide gas contains 1 mole of carbon and 2 moles of oxygen atoms.
According to mole concept:
1 mole of a compound contains number of molecules
So, 0.0784 moles of carbon dioxide gas will contain number of carbon atoms and
number of oxygen atoms
Hence, the number of carbon and oxygen atoms in the given amount of carbon dioxide is and
respectively