
Рассчитайте объем водорода,выделившегося при взаимодействии 30г цинка с 200 мл 20%-ного раствора
соляной кислоты

Ответы на вопрос

0.2=х:200
х=40
zn+2HCl=zncl2+h2
дальше через избыток недостаток
за х принимаем массу цинка
х:65=40:36,5
х=67,6 а у нас по условию 30, значит цинк в недостатке. расчёт ведем по недостатку. на у принимаем водород
30:65=у:22,4
у=10,3



Calculation of the Volume of Hydrogen Gas Released
To calculate the volume of hydrogen gas released when 30g of zinc reacts with 200ml of a 20% solution of hydrochloric acid, we need to use the balanced chemical equation for the reaction between zinc and hydrochloric acid:
Zn + 2HCl -> ZnCl2 + H2
From the equation, we can see that 1 mole of zinc reacts with 2 moles of hydrochloric acid to produce 1 mole of hydrogen gas. To find the volume of hydrogen gas, we need to convert the mass of zinc to moles and then use the mole ratio to determine the number of moles of hydrogen gas produced.
Step 1: Convert the mass of zinc to moles
The molar mass of zinc (Zn) is 65.38 g/mol. To convert the mass of zinc to moles, we divide the given mass by the molar mass:30g / 65.38 g/mol = 0.458 mol (approximately)
Step 2: Use the mole ratio to determine the moles of hydrogen gas
According to the balanced chemical equation, 1 mole of zinc reacts with 1 mole of hydrogen gas. Therefore, the number of moles of hydrogen gas produced is also 0.458 mol.Step 3: Convert moles of hydrogen gas to volume
To convert moles of hydrogen gas to volume, we need to use the ideal gas law, which states that:PV = nRT
Where: P = pressure (in atm) V = volume (in liters) n = number of moles R = ideal gas constant (0.0821 L·atm/mol·K) T = temperature (in Kelvin)
In this case, we are given the volume of the hydrochloric acid solution (200ml) and the concentration of the solution (20%). However, we do not have the pressure or temperature information. Therefore, we cannot calculate the exact volume of hydrogen gas released without additional information.
Please provide the pressure and temperature conditions for a more accurate calculation.


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