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.
Answer:
The number of mol is: 0, 042 mol in 4 grams of MgCl2
Explanation:
We calculate the weight of 1 mol of MgCl2:
Weight 1mol of MgCl2= weight Mg + (weight Cl)x 2=
24, 3 grams + 2 x 35, 5 grams = 95, 3 grams/mol MgCl2
95, 3 grams------1 mol MgCl2
4 grams -------x = (4 grams x1 mol MgCl2)/ 95, 3 grams= 0, 04197 mol MgCl2
Answer: You forgot to zero the balance
Explanation:
Hey there!
Balance the equation:
SiCl₄ + H₂O → H₄SiO₄ + HCl
Balance H.
2 on the left, 5 on the right. Add a coefficient of 3 in front of H₂O and a coefficient of 2 in front of HCl.
SiCl₄ + 3H₂O → H₄SiO₄ + 2HCl
Balance O.
3 on the left, 4 on the right. Change the coefficient of 3 in front of H₂O to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 2HCl
This unbalanced our H, so change the coefficient of 2 in front of HCl to a 4.
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Balance Cl.
4 on the left, 4 on the right. Already balanced.
Balance Si.
1 on the left, 1 on the right. Already balanced.
Our final balanced equation:
SiCl₄ + 4H₂O → H₄SiO₄ + 4HCl
Hope this helps!