Aluminum oxide produced : = 79.152 g
<h3>Further explanation</h3>
Given
46.5g of Al
165.37g of MnO
Required
Aluminum oxide produced
Solution
Reaction
2 Al (s) + 3 MnO (s) → 3 Mn (s) + Al₂O₃ (s)
mol = mass : Ar
mol = 46.5 : 27
mol = 1.722
mol = 165.37 : 71
mol = 2.329
mol : coefficient ratio Al : MnO = 1.722/2 : 2.329/3 = 0.861 : 0.776
MnO as a limiting reactant(smaller ratio)
So mol Al₂O₃ based on MnO as a limiting reactant
From equation , mol Al₂O₃ :
= 1/3 x mol MnO
= 1/3 x 2.329
= 0.776
Mass Al₂O₃ (MW=102 g/mol) :
= 0.776 x 102
= 79.152 g
Answer: 24 gram is a final Answer.
Explanation:1 mol of carbon weight is 12 gram then
12.044×1023 contain 2 mole carbon so
Answer:
See explanation
Explanation:
Molar mass of NaCl = 58.5 g
Number of moles contained in 10 g of NaCl = 10 g/58.5 g = 0.17 moles
If 1 mole of NaCl contains 6.02 * 10^23 atoms
0.17 moles of NaCl contains 0.17 * 6.02 * 10^23 atoms = 1.02 * 10^23 atoms
Molar mass of Fe II chloride = 126.751 g/mol
Number of moles = 10 g/126.751 g/mol = 0.0789 moles
Number of atoms = 0.0789 moles * 6.02 * 10^23 atoms = 4.7 * 10^22 atoms
Molar mass of Na = 23 g/mol
Number of moles = 10g/23 g/mol = 0.43 moles
Number of atoms = 0.43 moles * 6.02 * 10^23 atoms = 2.59 * 10^ 23 atoms
Answer:

Explanation:
(a) Litres of water

(b) Mass of NaF

1 kmol of NaF (41.99 kg) contains 19.00 kg of F⁻.

Answer:
See below.
Explanation:
The mass of octane in the sample of gasoline is 0.02851 * 482.6 = 13.759 g of octane.
The balanced equation is:
2C8H18(l) + 25O2(g) ----> 16CO2(g) + 18H2O(g)
From the equation, using atomic masses:
228.29 g of octane forms 704 g of CO2 and 324.3 g of H2O
So the mass of CO2 formed from the combustion of 13.759 g of octane = (704 * 13.759) / 228.29
= 42.43 g of CO2.
Amount of water = 324.3 * 13.759) / 228.29
= 19.55 g of H2O.