Question

Сколько граммов ZnCL2 и какой объем H2 образуется при взаимодействии Zn массой 650 граммов с HCL

Answer 1

Там нету грамов обьема

Answer 2

Reaction Equation and Stoichiometry

When zinc (Zn) with a mass of 650 grams reacts with hydrochloric acid (HCl), the following reaction occurs: Zn + 2HCl → ZnCl2 + H2

According to the stoichiometry of the reaction, 1 mole of Zn reacts with 2 moles of HCl to produce 1 mole of ZnCl2 and 1 mole of H2.

Calculation

To find the amount of ZnCl2 and the volume of H2 produced, we can use the following steps:

1. Calculate the moles of Zn using its molar mass. 2. Use the mole ratio from the balanced chemical equation to find the moles of ZnCl2 and H2. 3. Convert the moles of H2 to volume using the ideal gas law.

Step 1: Calculate Moles of Zn

The molar mass of Zn is 65.38 g/mol. Moles of Zn = Mass of Zn / Molar mass of Zn Moles of Zn = 650 g / 65.38 g/mol Moles of Zn ≈ 9.94 moles

Step 2: Use Mole Ratio to Find Moles of ZnCl2 and H2

From the balanced chemical equation: 1 mole of Zn produces 1 mole of ZnCl2 and 1 mole of H2.

So, moles of ZnCl2 = Moles of Zn Moles of ZnCl2 ≈ 9.94 moles

Moles of H2 = Moles of Zn Moles of H2 ≈ 9.94 moles

Step 3: Convert Moles of H2 to Volume

The ideal gas law is PV = nRT, where: P = pressure V = volume n = moles R = ideal gas constant T = temperature

Assuming standard temperature and pressure (STP), where T = 273 K and P = 1 atm, we can use the molar volume of a gas at STP, which is 22.4 L/mol.

Volume of H2 = Moles of H2 * Molar volume at STP Volume of H2 ≈ 9.94 moles * 22.4 L/mol Volume of H2 ≈ 223.1 L

Conclusion

When 650 grams of Zn reacts with HCl, approximately 9.94 moles of ZnCl2 are produced, and the volume of H2 formed is approximately 223.1 liters.