Question

Сколько граммов аммиачного раствора гидроксида серебра вступило в реакцию с этаналем если образовалось 10,8 г. серебра??

Answer 1

Найдем теоретический выход уксусной кислоты:64*100/80=80 гНайдем кол-во уксусной кислоты:n(CH3COOH)=m/M=80/60=1.3333 мольНайдем массу альдегида:m(CH3CHO)=M*n=1.3333*44=58.67 г

Answer 2

Calculating the Amount of Silver Hydroxide Solution Reacted with Ethanal

To calculate the amount of silver hydroxide solution that reacted with ethanal, we can use the given information that 10.8 grams of silver was formed in the reaction.

The balanced chemical equation for the reaction between ethanal and silver hydroxide is: AgOH + CH3CHO → Ag + CH3COOH + H2O

From the equation, we can see that 1 mole of AgOH reacts with 1 mole of CH3CHO to produce 1 mole of Ag.

The molar mass of Ag is 107.87 g/mol, and the molar mass of CH3CHO (ethanal) is 44.05 g/mol.

Using this information, we can calculate the amount of AgOH that reacted with CH3CHO.

Calculation

1. Calculate the moles of Ag formed: - Moles of Ag = Given mass of Ag / Molar mass of Ag - Moles of Ag = 10.8 g / 107.87 g/mol

2. Calculate the moles of AgOH reacted: - Moles of AgOH = Moles of Ag (from the balanced chemical equation)

3. Convert moles of AgOH to grams: - Grams of AgOH = Moles of AgOH * Molar mass of AgOH

Let's perform the calculations:

1. Moles of Ag: - Moles of Ag = 10.8 g / 107.87 g/mol = 0.1 moles

2. Moles of AgOH: - Moles of AgOH = 0.1 moles

3. Grams of AgOH: - Grams of AgOH = 0.1 moles * (107.87 g/mol + 16.00 g/mol + 1.01 g/mol) = 12.888 g

Answer:

The amount of silver hydroxide solution that reacted with ethanal is approximately 12.888 grams.