N₂ + 3H₂ ⇒ 2NH₃
1mol : 2mol
3,72mol : 7,44mol
n = 7,44mol
M = 17g/mol
m = n * M = 7,44mol * 17g/mol = 126,48g
Answer: fat show the first half and a lot of people in a row in this rr
Question 5 is the second one.
Answer:
3.3167 moles Of AlCl3
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
3Ca + 2AlCl3 —> 3CaCl2 + 2Al
From the balanced equation above,
2 moles of AlCl3 reacted to produce 2 moles of Al.
Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:
From the balanced equation above,
2 moles of AlCl3 reacted to produce 2 moles of Al.
Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.
Thus, 3.3167 moles Of AlCl3 is needed for the reaction.
To determine the amount of a certain element in a compound, we use the ratio of the elements from the compound. We calculate is follows:
45.0 g CCl4 ( 1 mol CCl4 / 153.82 g CCl4 ) ( 1 mol C / 1 mol CCl4 ) ( 12.01 g C / 1 mol C ) = 3.5135 g carbon present
Hope this answers the question. Have a nice day.
