Question

Какой объем водорода выделится при взаимодействии 49 г фосфорной кислоты с натрием, если выход продукта составляет 75%

Answer 1

1) 49  —  х л.
   196  —  67,2 л.  ; х=16, 8 л.

2) 16,8 л.-  100%
     х л.  —  75%    ;  х=12, 6 л.
   Ответ: 12, 6 л.

Answer 2

Calculation of Hydrogen Gas Volume

To calculate the volume of hydrogen gas released during the reaction between 49 g of phosphoric acid (H₃PO₄) and sodium (Na), we need to consider the balanced chemical equation for the reaction:

2 H₃PO₄ + 6 Na → 3 Na₂HPO₄ + 3 H₂O + 2 H₂

From the balanced equation, we can see that 2 moles of phosphoric acid (H₃PO₄) react to produce 2 moles of hydrogen gas (H₂). Therefore, we need to convert the given mass of phosphoric acid to moles and then use the mole ratio to determine the moles of hydrogen gas produced.

Step 1: Convert mass of phosphoric acid to moles

The molar mass of phosphoric acid (H₃PO₄) can be calculated by summing the atomic masses of its constituent elements:

H: 1.01 g/mol x 3 = 3.03 g/mol P: 30.97 g/mol O: 16.00 g/mol x 4 = 64.00 g/mol

Total molar mass of H₃PO₄ = 3.03 g/mol + 30.97 g/mol + 64.00 g/mol = 98.00 g/mol

To convert the mass of phosphoric acid to moles, we divide the given mass by the molar mass:

49 g / 98.00 g/mol = 0.5 mol

Therefore, we have 0.5 moles of phosphoric acid.

Step 2: Determine moles of hydrogen gas produced

According to the balanced equation, 2 moles of phosphoric acid (H₃PO₄) react to produce 2 moles of hydrogen gas (H₂). Therefore, the moles of hydrogen gas produced will be the same as the moles of phosphoric acid used.

Moles of hydrogen gas = 0.5 mol

Step 3: Calculate the volume of hydrogen gas

To calculate the volume of hydrogen gas, we need to use the ideal gas law equation:

PV = nRT

Where: P = pressure (assumed to be constant) V = volume of gas n = moles of gas R = ideal gas constant T = temperature (assumed to be constant)

Since the pressure and temperature are assumed to be constant, we can simplify the equation to:

V = nR

The ideal gas constant (R) is approximately 0.0821 L·atm/(mol·K).

Substituting the values, we have:

V = 0.5 mol x 0.0821 L·atm/(mol·K) = 0.04105 L

Therefore, the volume of hydrogen gas released during the reaction is approximately 0.04105 liters.

Please note that this calculation assumes ideal gas behavior and standard temperature and pressure conditions.