# Aluminum hydroxide reacts with nitric acid to form aluminum nitrate and water. What mass of water ca…

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### Tanong

Aluminum hydroxide reacts with nitric acid to form aluminum nitrate and water. What mass of water can be formed by the reaction of 15.0 g of aluminum hydroxide with excess nitric acid?

### Mga sagot sa #1 sa Tanong: Aluminum hydroxide reacts with nitric acid to form aluminum nitrate and water. What mass of water can be formed by the reaction of 15.0 g of aluminum hydroxide with excess nitric acid?

To **find the mass of water that formed from the reaction of 15.0g of aluminum hydroxide with nitric acid**, we should first do the following steps:

**Write the aluminum hydroxide and nitric acid net equation**

**Balanced equation of the aluminum hydroxide + nitric acid**

- Find for the molar mass of aluminum hydroxide

- Find for the molar mass of water

After finding all of these, we can then solve for the mass of water that is formed from the reaction of 15.0 g of aluminum hydroxide.

Write the aluminum hydroxide and nitric acid net equation

[tex]Al(OH)_{3}[/tex] + [tex]HNO_{3}[/tex] → [tex]Al(NO_{3} )_{3}[/tex] + [tex]H_{2}O[/tex]

**Write a balanced equation for the reaction of aluminum hydroxide + nitric acid
**

[tex]Al(OH)_{3}[/tex] + 3 [tex]HNO_{3}[/tex] → [tex]Al(NO_{3} )_{3}[/tex] + 3 [tex]H_{2}O[/tex]

Find for the molar mass of aluminum hydroxide

Al - 1 x 26.98 = 26.98

O - 3 x 16 = 48

H - 3 x 1.01 = 3.03

Total = 78.01 g/mol

Find for the molar mass of water

H - 2 x 1.01 = 2.02

O - 1 x 16 = 16

Total = 18.02 g /mol

Let us solve for the mass of water.

15.0 g [tex]Al(OH)_{3}[/tex] x[tex]frac{1 mol Al(OH)_{3}}{78.01 g / mol Al(OH)_{3}}[/tex] x [tex]frac{3 mol H_{2}O}{1 mol Al(OH)_{3}}[/tex] x [tex]frac{18.02 g H_{2}O }{1 mol H_{2}O}[/tex]

= [tex]frac{810.09 mol H_{2}O}{78.01}[/tex] = 10.39 or 10.4 g [tex]H_{2}O[/tex]

Thus, the mass of water that formed from the reaction of 15.0g of aluminum hydroxide with nitric acid is 10.4 g.

Other stoichiometric related are the following:

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