M (zn)=6.5 грамм
w (ha2so4)=19.6%
v (h2)=?

Предыдущий
вопрос Следующий
вопрос

To find the volume of hydrogen gas produced in the reaction, we use the stoichiometry of the reaction and the given masses of zinc and sulfuric acid.

The balanced chemical equation for the reaction between zinc and sulfuric acid is:

Zn + H2SO4 -> ZnSO4 + H2

From the equation, we see that 1 mole of zinc reacts with 1 mole of sulfuric acid to produce 1 mole of hydrogen gas.

First, we calculate the number of moles of zinc present:

molar mass of Zn = 65.38 g/mol
moles of Zn = mass/molar mass = 6.5 g / 65.38 g/mol = 0.0994 mol

Next, we calculate the number of moles of H2SO4 present:
mass of H2SO4 = 19.6% of total mass = 19.6/100 * 6.5 g = 1.274 g
molar mass of H2SO4 = 98.08 g/mol
moles of H2SO4 = mass/molar mass = 1.274 g / 98.08 g/mol = 0.013 mol

Since the reaction between Zn and H2SO4 occurs in a 1:1 mole ratio, the moles of hydrogen gas produced is equal to the moles of H2SO4. Therefore, the moles of H2 produced is 0.013 mol.

Finally, we calculate the volume of hydrogen gas at STP (standard temperature and pressure, 0°C and 1 atm) using the ideal gas law:

PV = nRT

where P = pressure (1 atm), V = volume, n = moles of gas, R = ideal gas constant (0.0821 L.atm/mol.K), and T = temperature (273 K)

V = nRT/P = 0.013 mol 0.0821 L.atm/mol.K 273 K / 1 atm = 0.292 L

Therefore, the volume of hydrogen gas produced in the reaction is 0.292 liters.