Question

СРОЧНО ДАЮ 50 БАЛЛОВ СРОЧНО СРОЧНО СРОЧНО Укажіть масу барій гідроксиду (в грамах), що витратиться на нейтралізацію розчину, де міститься 12,6 г нітратної кислоти. Відповідь надайте з точністю до десятих.​

Answer 1

Ответ:

17,1 г

7Объяснение:

Answer 2

Ba(OH)2 + 2HNO3 = Ba(NO3)2 + 2H2O

m(HNO3) = 12,6 г

m(Ba(OH)2) - ?

Решение:

1) μ(HNO3) = 12,6 г / 63 г/моль = 0,2 моль

2) μ(Ba(OH)2) = 0,2 моль / 2 = 0,1 моль

3) m(Ba(OH)2) = 0,1 моль * 171 г/моль = 17,1 г

Ответ: 17,1 г

Answer 3

To find the mass of barium hydroxide required to neutralize a solution containing 12.6 g of nitric acid, we need to consider the balanced chemical equation for the reaction between barium hydroxide (Ba(OH)2) and nitric acid (HNO3):

2 Ba(OH)2 + 2 HNO3 -> Ba(NO3)2 + 2 H2O

From the balanced equation, we can see that 2 moles of barium hydroxide react with 2 moles of nitric acid to produce 1 mole of barium nitrate and 2 moles of water.

To determine the mass of barium hydroxide, we need to calculate the number of moles of nitric acid present in the solution. We can use the molar mass of nitric acid (HNO3) to convert the given mass of nitric acid to moles:

Molar mass of HNO3 = 1(1.01) + 1(14.01) + 3(16.00) = 63.01 g/mol

Number of moles of HNO3 = mass of HNO3 / molar mass of HNO3 = 12.6 g / 63.01 g/mol = 0.20 mol

Since the stoichiometric ratio between nitric acid and barium hydroxide is 1:1, we can conclude that 0.20 moles of barium hydroxide are required to neutralize the given amount of nitric acid.

Now, we can calculate the mass of barium hydroxide using its molar mass:

Molar mass of Ba(OH)2 = 1(137.33) + 2(1.01) + 2(16.00) = 171.34 g/mol

Mass of Ba(OH)2 = number of moles of Ba(OH)2 * molar mass of Ba(OH)2 = 0.20 mol * 171.34 g/mol = 34.27 g

Therefore, the mass of barium hydroxide required to neutralize the solution containing 12.6 g of nitric acid is 34.27 grams [[1]].