Question

Какая масса пропанола вступила в реакцию с избытком металлического калия если в ходе реакцию выделялся газ объёмом 560 см3 н. у?

Answer 1

2C3H7OH+2K=2C3H7OK+H2 
Определим что в недостатки ,по нему будим вести расчет 
nC3H7OH=7,2/Mr(36+7+17=60)=0,12 моль
nK=5,85/39=0,15 моль
0,12<0,15
nC3H7OK=nC3H7OH=0,12 моль
m---------=0,12*Mr(36+7+16+39=98)=11,76г 
2) 
2С3Н7ОН+2Na=2C3H7ONa+H2 
nH=14/22,4=0,625 моль
nC3H7OH=1/2nH=0,625/2=0,313 моль
mс3Н7ОН=0,313*60=18,75г

Answer 2

Reaction of Propanol with Excess Metallic Potassium

To calculate the mass of propanol that reacted with excess metallic potassium, we can use the information provided about the volume of gas produced and the reaction equation.

The reaction of propanol with metallic potassium can be represented by the following equation: ``` C3H7OH + K → C3H7OK + 1/2H2 ```

Calculation

Given: - Volume of gas produced (V) = 560 cm^3 at standard conditions - The molar volume of gas at standard conditions is approximately 22.4 L/mol

Using the ideal gas law, we can calculate the number of moles of gas produced: ``` n = V / Vm = 560 cm^3 / 22.4 L/mol ≈ 0.025 mol ```

From the balanced chemical equation, we know that 1 mole of propanol reacts to produce 1/2 mole of hydrogen gas. Therefore, the number of moles of propanol that reacted is twice the number of moles of hydrogen gas produced: ``` n(propanol) = 2 * n(H2) = 2 * 0.025 mol = 0.05 mol ```

Molar Mass of Propanol

The molar mass of propanol (C3H7OH) can be calculated as follows: ``` Molar mass = (3 * molar mass of carbon) + (8 * molar mass of hydrogen) + molar mass of oxygen = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) + 16.00 g/mol = 60.15 g/mol ```

Mass of Propanol

Finally, we can calculate the mass of propanol that reacted: ``` Mass = n * molar mass = 0.05 mol * 60.15 g/mol ≈ 3.008 g ```

The mass of propanol that reacted with excess metallic potassium is approximately 3.008 grams.