What volume, in liters, of 0.500 M HNO3 is necessary to react with 25.0 mL of 0.0500 M Ca(OH)2 solution to the endpoint? HNO3 + Ca(OH)2. --> Ca(NO3)2 + H2O

0.00500L are necessaries

Explanation:

Based on the balanced chemical reaction:

2HNO3 + Ca(OH)2. --> Ca(NO3)2 + 2H2O

2 moles of nitric acid react with 1 mole of calcium hydroxide

To solve this question we must find the moles of Ca(OH)2. Then, convert this moles to moles of HNO3 using the chemical reaction and to liters:

Moles Ca(OH)2:

0.025L * (0.0500mol / L) = 0.00125 moles Ca(OH)2

Moles HNO3:

0.00125 moles Ca(OH)2 * (2 moles HNO3 / 1mol Ca(OH)2) = 0.0025 moles HNO3

Liters HNO3:

0.0025 moles HNO3 * (1L / 0.500moles) = 0.00500L are necessaries

5.6⋅10−^19  j

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