Answer:
2.4 moles of oxygen are needed to react with 87 g of aluminium.
Explanation:
Chemical equation:
4Al(s) + 3O₂(l) → 2AlO₃(s)
Given data:
Mass of aluminium = 87 g
Moles of oxygen needed = ?
Solution:
Moles of aluminium:
Number of moles of aluminium= Mass/ molar mass
Number of moles of aluminium= 87 g/ 27 g/mol
Number of moles of aluminium= 3.2 mol
Now we will compare the moles of aluminium with oxygen.
Al : O₂
4 : 3
3.2 : 3/4×3.2 = 2.4 mol
2.4 moles of oxygen are needed to react with 87 g of aluminium.
The naturally acidic rainwater will erode the limestone mountin over time
Answer:
The chemist would require to use 43.43 grams.
Explanation:
In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:
- 0.550 mol * 78.96 g/mol = 43.43 g
The chemist would require to use 43.43 grams.
Should be 1.8L.
2 moles of hydrogen react with 1 mole of oxygen. If 2 moles of hydrogen is 3.6L, 1 mole of oxygen should be 1.8L.
Answer: 2800 g
Explanation:
According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number 
of particles.
Given mass = 5 kg = 5000 g
1 mole of 
produces = 1 mole of 
50 moles of produces =
of
Mass of 
2800 g of is produced from 5.0 kg of limestone.
