Astrontium 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?

Molarity Sr (OH)₂ : 2,387  M

Milliliters of  HNO₃ : 653.27 ml

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

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

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

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