How many moles of Al are needed to form 2.43 mol of Al2Br6?
(2.43 mol Al2Br6)*(2 mol Al/ 1 mol Al2Br6)= 4.86 mol Al
Well, I'm not exactly a fan of moles, but I'll give it a shot anyway! So, to form Al2Br6, you'll need twice the number of moles of aluminum as the number of moles of Al2Br6. Since you're starting with 2.43 moles of Al2Br6, you'll need 2.43 multiplied by 2 moles of Al. That's a grand total of 4.86 moles of Al! That's a lot of moles, but hey, who's counting? Well, I guess we are in this case. Just remember, moles may be small, but they can sure make a big mess in your chemistry lab!
To determine how many moles of Al are needed to form 2.43 mol of Al2Br6, we can refer to the balanced chemical equation for the formation of Al2Br6.
The balanced equation is:
2 Al + 3 Br2 -> 2 Al2Br6
According to the equation, 2 moles of Al are needed to produce 2 moles of Al2Br6. Therefore, we can set up a ratio:
2 moles Al / 2 moles Al2Br6 = x moles Al / 2.43 moles Al2Br6
By cross-multiplying and solving for x, we can determine the number of moles of Al needed:
2 moles Al * 2.43 moles Al2Br6 = 2 moles Al2Br6 * x moles Al
4.86 moles Al = 2 moles Al2Br6 * x
Dividing both sides by 2 moles Al2Br6:
x = 4.86 moles Al / 2 moles Al2Br6
x ≈ 2.43 moles Al
Therefore, approximately 2.43 moles of Al are needed to form 2.43 moles of Al2Br6.
To determine how many moles of Al are needed to form 2.43 mol of Al2Br6, we need to examine the chemical formula of Al2Br6.
The formula tells us that there are two moles of aluminum (Al) for every one mole of Al2Br6. Therefore, the number of moles of Al present in 2.43 mol of Al2Br6 can be calculated as follows:
Number of moles of Al = (Number of moles of Al2Br6) × (Number of moles of Al / Number of moles of Al2Br6)
Number of moles of Al = 2.43 mol × (2 mol Al / 1 mol Al2Br6)
Number of moles of Al = 4.86 mol Al
So, to form 2.43 mol of Al2Br6, you would need 4.86 moles of Al.