Answer:
There are <u>4</u><u>3</u><u>4</u><u>7</u><u> </u>alkanes with that molecular formula.
I hope this helps
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:
Since there are 3 Hydrogen atoms present, the formula mass of H is 1.0 × 3 = 3.0 g/mol. Therefore, by adding them up, the formula mass of ammonia is: [14.0 g/mol + 3.0 g/mol] = 17.0 g/mol.
The significant figures are 2 and 5 in 250.
there are 1023 atoms in pure aluminium
