A strontium hydroxide solution is prepared by dissolving 21.78 g of strontium hydroxide in water to make 75.00 mL of solution. What is the molarity of this solution? Next the strontium hydroxide solution prepared in part A is used to react with a 0.250 M nitric acid solution. Write a balanced chemical equation to represent the reaction between the strontium hydroxide and nitric acid solutions. How many milliliters of the nitric acid solution from part B (0.250 M) is needed to completely react 34.21 mL of the strontium hydroxide solution from part A?

Question

Chemistry

Kamari Brock

23 October, 13:31

A strontium hydroxide solution is prepared by dissolving 21.78 g of strontium hydroxide in water to make 75.00 mL of solution. What is the molarity of this solution? Next the strontium hydroxide solution prepared in part A is used to react with a 0.250 M nitric acid solution. Write a balanced chemical equation to represent the reaction between the strontium hydroxide and nitric acid solutions. How many milliliters of the nitric acid solution from part B (0.250 M) is needed to completely react 34.21 mL of the strontium hydroxide solution from part A?

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Allison Richard · Answer 1

Chemical reaction: Sr (OH) ₂ + 2HNO₃ → Sr (NO₃) ₂ + 2H₂O.

part A) m (Sr (OH) ₂) = 21,78 g.

n (Sr (OH) ₂) = m (Sr (OH) ₂) : M (Sr (OH) ₂).

n (Sr (OH) ₂) = 21,78 g : 87,62 g/mol.

n (Sr (OH) ₂) = 0,248 mol.

c (Sr (OH) ₂) = n (Sr (OH) ₂) : V (Sr (OH) ₂).

c (Sr (OH) ₂) = 0,248 mol : 0,075 L.

c (Sr (OH) ₂) = 3,3 mol/L.

part B) V (Sr (OH) ₂) = 34,21 mL = 0,03421 L.

n (Sr (OH) ₂) = 0,03421 L · 3,3 mol/L = 0,112 mol.

From chemical reaction: n (Sr (OH) ₂) : n (HNO₃) = 1 : 2.

n (HNO₃) = 0,225 mol.

V (HNO₃) = 0,225 mol : 0,25 mol/L.

V (HNO₃) = 0,9 L = 900 ml.