Question

Какой объем водорода выделится при взаимодействии 13г цинка с 7.3г 10% соляной кислоты

Answer 1

Zn+2HCl=ZnCl2+H2
находим массу соляной кислоты=7,3х0,1=0,73 г
находим число молей соляной кислоты=0,73\36,5=0,02 моль
число молей цинка=13\65=0,2 моль
получатся цинк взят в избытке
определяем объём водорода по соляной кислоте
число молей водорода=0,02\2=0,01 моль
объём водорода=0,01х22,4=0,224 литра

Answer 2

Reaction of Zinc with Hydrochloric Acid

When 13g of zinc reacts with 7.3g of 10% hydrochloric acid, we can calculate the volume of hydrogen gas produced using the given information.

Calculation

First, we need to find the limiting reactant to determine the amount of hydrogen gas produced. The balanced chemical equation for the reaction is:

Zn + 2HCl → ZnCl2 + H2

From the given information, we can calculate the moles of each reactant using their molar masses and then use the stoichiometry of the reaction to find the amount of hydrogen gas produced.

1. Molar Mass of Zinc (Zn): - Molar mass of Zn = 65.38 g/mol.

2. Molar Mass of Hydrochloric Acid (HCl): - Molar mass of HCl = 36.46 g/mol.

3. Calculation of Moles: - Moles of Zn = Mass of Zn / Molar mass of Zn - Moles of Zn = 13g / 65.38 g/mol - Moles of Zn = 0.1987 mol - Moles of HCl = (Mass of HCl * % concentration) / Molar mass of HCl - Moles of HCl = (7.3g * 0.10) / 36.46 g/mol - Moles of HCl = 0.0200 mol

4. Limiting Reactant: - The stoichiometry of the reaction shows that 1 mole of Zn reacts with 2 moles of HCl. Therefore, the limiting reactant is the one that produces the least amount of product. - Since the moles of Zn (0.1987 mol) are greater than twice the moles of HCl (0.0200 mol), HCl is the limiting reactant.

5. Calculation of Hydrogen Gas: - Using the stoichiometry of the reaction, 2 moles of HCl produce 1 mole of H2. - Moles of H2 produced = 0.0200 mol * (1/2) = 0.0100 mol

6. Volume of Hydrogen Gas: - The volume of 1 mole of any gas at standard temperature and pressure (STP) is 22.4 liters. - Volume of H2 produced = Moles of H2 * 22.4 L/mol - Volume of H2 produced = 0.0100 mol * 22.4 L/mol - Volume of H2 produced = 0.224 L

Conclusion

When 13g of zinc reacts with 7.3g of 10% hydrochloric acid, approximately 0.224 liters of hydrogen gas will be produced.

This calculation is based on the stoichiometry of the reaction and the given masses of the reactants.