Question

Plz help asap. answer needed!!!

Answer 1

Answer:

11.43g of Aluminum Hydroxide

Explanation:

Since we know that the sulfuric acid is the limiting reactant in this chemical reaction, we know that we are going to be left with excess aluminum hydroxide. So to find the amount of leftover aluminum hydroxide we are going to need to convert the given amount of sulfuric acid to the amount of aluminum hydroxide needed to react with the sulfuric acid.

$\frac{35g H_{2}SO_{4}}{1}*\frac{1 mole H_{2}SO_{4}}{98.079 g H_{2}SO_{4}} *\frac{2 moles Al(OH)_{3} }{3 moles H_{2}SO_{4}} * \frac{78.003 g Al(OH)_{3} }{1 mole Al(OH)_{3} } = 18.557 g Al(OH)_{3}$

Once you do that, you need to subtract that number from the amount of aluminum hydroxide given to get the amount of left over aluminum hydroxide.

$30 g Al(OH)_{3} - 18.557 g Al(OH)_{3} = 11.43 g Al(OH)_{3}$

Hope this helps!