Answer:
0.7g of HCl
Explanation:
First, let us write a balanced equation for the reaction between HCl and Al(OH)3.
This is illustrated below:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:
Molar Mass of Al(OH)3 = 27 + 3(16+1)
= 27 + 3(17) = 27 + 51 = 78g/mol.
Molar Mass of HCl = 1 + 35.5 = 36.5g/mol
Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g
Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:
Al(OH)3 + 3HCl —> AlCl3 + 3H2O
From the equation above,
78g of Al(OH)3 reacted with 109.5g of HCl.
Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl
Answer:
C6H14O3F
Explanation:
The first step is to divide each compound by its molecular weight
Carbon
= 39.10/12
= 3.258
Hydrogen
= 7.67/1
= 7.67
Oxygen
= 26.11/16
= 1.63
Phosphorous
= 16.82/31
= 0.542
Flourine
= 10.30/19
= 0.542
The next step is to divide by the lowes value
3.258/0.542
= 6 mol of C
7.67/0.542
= 14 mol of H
1.63/0.542
= 3 mol of O
0.542/0.542
= 1 mol of P
0.542/0.542
= 1 mol of F
Hence the molecular formula is C6H14O3F
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