Aluminum bromide and sodium hydroxide react to form aluminum hydroxide and sodium bromide.
a.)How many moles of sodium bromide can be formed from 1.55 moles of aluminum bromide?
b.)How many moles of aluminum hydroxide may be formed from 4.65 moles of sodium hydroxide?
AlBr3 + 3NaOH ==> Al(OH)3 + 3NaBr
1.55 mols AlBr3 x (32 mols NaBr/1 mol AlBr3) = 1.55 x 3/1 = ?
You do part b the same way. Use the coefficients in the balanced equation to convert what you have to what you want; i.e., to convert anything to anything.
Aluminum bromide, Al(Br)3, and sodium hydroxide, NaOH, react to form aluminum hydroxide, Al(OH)3, and sodium bromide, NaBr.
a) How many moles of sodium bromide can be formed from 1.55 moles of aluminum bromide?
a.) Well, it's bromide-ing that you asked! Let's use the balanced equation to figure this out. One mole of aluminum bromide reacts with three moles of sodium hydroxide to produce three moles of sodium bromide. So, if we have 1.55 moles of aluminum bromide, we can expect to form 1.55 * 3 = 4.65 moles of sodium bromide. That's a whole lot of bromide goodness!
b.) Oh, hydroxide-y dokey! According to the balanced equation, one mole of sodium hydroxide reacts with one mole of aluminum hydroxide. So, if we have 4.65 moles of sodium hydroxide, we can expect to form 4.65 moles of aluminum hydroxide. That's a mole lot of aluminum hydroxide, don't you think?
To answer these questions, we need to use the balanced chemical equation for the reaction:
Aluminum bromide + Sodium hydroxide → Aluminum hydroxide + Sodium bromide
The balanced equation shows that the stoichiometry of the reaction is 1:1. That means 1 mole of aluminum bromide reacts with 1 mole of sodium hydroxide to produce 1 mole of aluminum hydroxide and 1 mole of sodium bromide.
a.) To determine how many moles of sodium bromide can be formed from 1.55 moles of aluminum bromide, we can use the ratio from the balanced equation:
1 mole of aluminum bromide reacts with 1 mole of sodium bromide.
Therefore, if 1 mole of aluminum bromide produces 1 mole of sodium bromide, then 1.55 moles of aluminum bromide will also produce 1.55 moles of sodium bromide.
So, 1.55 moles of aluminum bromide can form 1.55 moles of sodium bromide.
b.) Similarly, to determine how many moles of aluminum hydroxide can be formed from 4.65 moles of sodium hydroxide, we can again use the ratio from the balanced equation:
1 mole of sodium hydroxide reacts with 1 mole of aluminum hydroxide.
Therefore, if 1 mole of sodium hydroxide produces 1 mole of aluminum hydroxide, then 4.65 moles of sodium hydroxide will also produce 4.65 moles of aluminum hydroxide.
So, 4.65 moles of sodium hydroxide can form 4.65 moles of aluminum hydroxide.
To answer these questions, we need to use the balanced chemical equation for the reaction between aluminum bromide (AlBr3) and sodium hydroxide (NaOH):
2AlBr3 + 6NaOH -> 2Al(OH)3 + 6NaBr
a.) How many moles of sodium bromide can be formed from 1.55 moles of aluminum bromide?
From the balanced equation, we can see that 2 moles of aluminum bromide react with 6 moles of sodium hydroxide to produce 6 moles of sodium bromide.
So, the mole ratio between aluminum bromide and sodium bromide is 2:6 or 1:3.
To find the number of moles of sodium bromide formed from 1.55 moles of aluminum bromide, we can set up a proportion:
1.55 moles AlBr3 / 2 moles AlBr3 = x moles NaBr / 6 moles NaBr
Cross-multiplying gives:
1.55 moles AlBr3 * 6 moles NaBr = 2 moles AlBr3 * x moles NaBr
9.3 moles NaBr = 2x moles NaBr
x = 9.3 moles NaBr / 2
x = 4.65 moles NaBr
So, from 1.55 moles of aluminum bromide, you can form 4.65 moles of sodium bromide.
b.) How many moles of aluminum hydroxide may be formed from 4.65 moles of sodium hydroxide?
From the balanced equation, we can see that 6 moles of sodium hydroxide react with 2 moles of aluminum bromide to produce 2 moles of aluminum hydroxide.
So, the mole ratio between sodium hydroxide and aluminum hydroxide is 6:2 or 3:1.
To find the number of moles of aluminum hydroxide formed from 4.65 moles of sodium hydroxide, we can set up a proportion:
4.65 moles NaOH / 6 moles NaOH = x moles Al(OH)3 / 2 moles Al(OH)3
Cross-multiplying gives:
4.65 moles NaOH * 2 moles Al(OH)3 = 6 moles NaOH * x moles Al(OH)3
9.3 moles Al(OH)3 = 6x moles Al(OH)3
x = 9.3 moles Al(OH)3 / 6
x = 1.55 moles Al(OH)3
So, from 4.65 moles of sodium hydroxide, you can form 1.55 moles of aluminum hydroxide.