Question

Какая масса соли образуется после взаимодействия 15гр 95%-ного раствора гидроксида бария с избытком соляной кислоты?

Answer 1

М в-ва(Ba(OH)20 = м р-ра *ω/100% = 15г*95%/100%=14,25г
14,25г                 хг
Ba(OH)2 +2HCl=BaCl2  + 2H2O
171г                   208г
х= 14,25*208/171=17,33г

Answer 2

Calculation of Salt Mass Formed

To calculate the mass of salt formed after the interaction of 15g of a 95% solution of barium hydroxide with an excess of hydrochloric acid, we can use the following steps:

1. Calculate the moles of barium hydroxide in the given solution. 2. Determine the limiting reactant between barium hydroxide and hydrochloric acid. 3. Use the stoichiometry of the balanced chemical equation to find the moles of salt formed. 4. Convert the moles of salt to grams.

Step 1: Calculate the Moles of Barium Hydroxide

The molarity of the 95% solution of barium hydroxide is not provided. However, we can assume it to be 95% w/v (weight/volume) solution. Using the density of barium hydroxide, we can calculate the molarity and then the moles of barium hydroxide.

Step 2: Determine the Limiting Reactant

Once we have the moles of barium hydroxide, we can compare it with the moles of hydrochloric acid to determine the limiting reactant.

Step 3: Use Stoichiometry

After identifying the limiting reactant, we can use the balanced chemical equation to find the moles of salt formed.

Step 4: Convert Moles to Grams

Finally, we can convert the moles of salt to grams using the molar mass of the salt.

Let's proceed with the calculations based on the provided information.

Step 1: Calculate the Moles of Barium Hydroxide The molarity of the 95% solution of barium hydroxide is not provided. Therefore, we cannot directly calculate the moles of barium hydroxide without the molarity or density information.

Conclusion: Without the molarity or density of the barium hydroxide solution, it is not possible to calculate the moles of barium hydroxide and proceed with the remaining steps of the calculation.