Question

Какой объём водорода выделится при соединении 260 грамм цинка с фосфорной кислотой.

Answer 1

Zn+ HPO4=ZnPO4+H2
ню(Zn)= 260г/65г/моль=4моль
ню(Н2)=4 моль
V(H2)=4моль*22,4л/моль= 89,6л

Answer 2

Calculation of Hydrogen Gas Produced

To calculate the volume of hydrogen gas produced when 260 grams of zinc reacts with phosphoric acid, we need to consider the balanced chemical equation for the reaction:

Zn + H3PO4 -> Zn3(PO4)2 + H2

From the balanced equation, we can see that 1 mole of zinc reacts with 2 moles of phosphoric acid to produce 1 mole of hydrogen gas. To determine the moles of zinc, we need to convert the given mass of zinc to moles using its molar mass.

The molar mass of zinc (Zn) is approximately 65.38 g/mol Therefore, the number of moles of zinc can be calculated as follows:

Moles of zinc = Mass of zinc / Molar mass of zinc = 260 g / 65.38 g/mol ≈ 3.97 mol

Since the stoichiometry of the reaction tells us that 1 mole of zinc produces 1 mole of hydrogen gas, we can conclude that 3.97 moles of zinc will produce 3.97 moles of hydrogen gas.

Now, to calculate the volume of hydrogen gas produced, we can use the ideal gas law equation:

PV = nRT

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

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

V = nR / P

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

Now, we need to determine the pressure of the gas. Unfortunately, the given information does not provide the pressure or temperature conditions under which the reaction occurs. Therefore, we cannot calculate the exact volume of hydrogen gas produced without this information.

If you have the pressure and temperature conditions, please provide them so that we can calculate the volume of hydrogen gas produced accurately.