Answer:
Explanation:
1) Chemical formula of sodium carbonate: <em>Na₂CO₃</em>
2) Ratio of carbon atoms:
- The number of atoms of C in the unit formula Na₂CO₃ is the subscript for the atom, which is 1 (since it is not written).
Hence, the ratio is 1 C atom / 1 Na₂CO₃ unit formula.
This is, there is 1 atom of carbon per each unit formula of sodium carbonate.
3) Calculate the number of moles in 1.773 × 10⁷ carbon atoms
- Divide by Avogadro's number: 6.022 × 10²³ atoms / mol
- number C moles = 1.773 × 10⁷ atoms / (6.022 × 10²³ atoms/mol)
- number C moles = 2.941 × 10⁻¹⁷ mol
Since, the ratio is 1: 1, the number of moles of sodium carbonate is the same number of moles of carbon atoms.
In order to get HgO you would need 2Hg+1O2=2HgO. Since oxygen is diatomic you need two when it stands alone causing you to need two mercuries to balance out the reactants and the product I hope this helps
Answer:
Explanation:
Hello,
In this case, for the given chemical reaction, we first identify the limiting reactant by noticing that due to the 1:1 mole ratio for magnesium to iodine the reacting moles must the same, nevertheless, there are only 2.68 moles of magnesium versus 3.56 moles of iodine, for that reason, magnesium is the limiting reactant, so the theoretical turns out:
Thus, we compute the percent yield as:
Best regards.
<span>If one of the impurities in diesel fuel has the formula c2h6s, then the products that will form would be carbon dioxide, water and sulfur dioxide. The balanced chemical reaction would be as follows:
</span>C2H6S(l)<span> + 9/2O2(g) = 2CO2(g) + 3H2O(v) + SO2(g)
</span>
Hope this answers the question.
Answer:
oxygen is limiting reactant
Explanation:
Given data:
Mass of hydrogen = 16.7 g
Mass of oxygen = 15.4 g
Limiting reactant = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 16.7 g/ 2 g/mol
Number of moles = 8.35 mol
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 15.4 g/ 32 g/mol
Number of moles = 0.48 mol
Now we will compare the moles of both reactant with product,
H₂ : H₂O
2 : 2
8.35 : 8.35
O₂ : H₂O
1 : 2
0.48 : 2×0.48 = 0.96 mol
The number of moles of water produced by oxygen are less so it will limiting reactant.
