Question

Обчисліть масу гліцеролу що прореагував з надлишком металічного натрію якщо під час реакції утворився водень об'ємом 13.44л(н.у)

Answer 1

2C3H5(OH)3+6Na=2C3H5(ONa)3+3H2

n(H2)=13.44/22.4=0.6 моль
2/3n(H2)=n(C3H5(OH)3)=0.4 моль
m(гліц.)=0.4 моль * 92 г/моль = 36.8 г

Answer 2

Calculating the Mass of Glycerol Reacted with Excess Metallic Sodium

To calculate the mass of glycerol that reacted with excess metallic sodium, we can use the stoichiometry of the reaction and the volume of hydrogen gas produced.

The balanced chemical equation for the reaction between glycerol and metallic sodium is:

C3H8O3 + 3Na → 3CH3OH + NaOH

From the balanced equation, we can see that 1 mole of glycerol reacts with 3 moles of metallic sodium to produce 3 moles of methanol.

Given that the volume of hydrogen gas produced during the reaction is 13.44 L at standard temperature and pressure (STP), we can use the relationship between volume, moles, and molar volume at STP to find the moles of hydrogen gas produced.

Using the molar volume of a gas at STP (22.4 L/mol), we can calculate the moles of hydrogen gas produced:

Moles of hydrogen gas = Volume of hydrogen gas / Molar volume at STP Moles of hydrogen gas = 13.44 L / 22.4 L/mol Moles of hydrogen gas ≈ 0.6 moles

Since 1 mole of glycerol produces 3 moles of hydrogen gas, the moles of glycerol reacted can be calculated as:

Moles of glycerol reacted = Moles of hydrogen gas / 3 Moles of glycerol reacted ≈ 0.6 moles / 3 Moles of glycerol reacted ≈ 0.2 moles

Now, to find the mass of glycerol that reacted, we can use the molar mass of glycerol, which is approximately 92.09 g/mol.

Mass of glycerol reacted = Moles of glycerol reacted * Molar mass of glycerol Mass of glycerol reacted ≈ 0.2 moles * 92.09 g/mol Mass of glycerol reacted ≈ 18.42 grams

Therefore, the mass of glycerol that reacted with excess metallic sodium is approximately 18.42 grams.