Question

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?

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.

Answer 2

Molarity Sr (OH)₂ : 2,387  M

Milliliters of  HNO₃ : 653.27 ml

Further explanation

Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products

• Part A

dissolving 21.78 g of strontium hydroxide in water to make 75.00 mL of solution, the molarity:

Molar mass of Sr (OH)₂ is 121.6347 g / mol e

mole Sr (OH)₂ = 21.78 g: 121.6347 = 0.179

M = 0.179 mole: 0.075 Liter = 2,387

• Part B

A balanced chemical equation

Sr (OH) ₂ + 2HNO₃ → Sr (NO₃) ₂ + 2H₂O

mole of 34.21 mL of Sr (OH) ₂ =

mole = 34.21 ml x 2,387 M

mole Sr (OH) ₂= 81,659 mlmole

from the chemical equation above Sr(OH)₂ mole : 2HNO₃ = 1: 2

so the mole needed by HNO₃ to react perfectly with Sr (OH) ₂ = 2 x 81,659 = 163,319 mlmole

So the amount of HNO₃ volume :

volume: 163,319 mmole x 0.25 M

volume = 653.27 ml

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Keywords: mole, strontium hydroxide solution,  a balanced chemical equation