Question

Смесь, состоящую из бария массой 13,7 г и оксида бария массой 15,3 г растворили в воде. Рассчитайте, гидроксид бария какой массы образовался при этом.

Answer 1

BaO + H2O = Ba(OH)2
Ba(OH)2 + CO2 = BaCO3 + H2O
n(BaO)=15,3/153=0,1 моль
n(BaCO3)=n(CO2)=n(Ba(OH)2)=n(BaO)=0,1 моль
V(CO2)=22,4*0,1=2,24 л
m(BaCO3)=197*0,1=19,7 г

Answer 2

О обязан отдых задних друзьях заряд

Answer 3

Calculating the Mass of Barium Hydroxide Formed

To calculate the mass of barium hydroxide formed when a mixture consisting of 13.7 g of barium and 15.3 g of barium oxide is dissolved in water, we can use the following chemical equation:

Ba + H2O → Ba(OH)2

First, we need to find the amount of barium oxide that reacts with water to form barium hydroxide. Then, we can calculate the mass of barium hydroxide formed.

Step 1: Calculate the Moles of Barium and Barium Oxide

The molar mass of barium (Ba) is 137.33 g/mol, and the molar mass of barium oxide (BaO) is 153.33 g/mol.

The number of moles of barium is calculated as: moles of Ba = mass of Ba / molar mass of Ba

The number of moles of barium oxide is calculated as: moles of BaO = mass of BaO / molar mass of BaO

Using the given masses: moles of Ba = 13.7 g / 137.33 g/mol moles of BaO = 15.3 g / 153.33 g/mol

Step 2: Determine the Limiting Reactant

The limiting reactant is the one that is completely consumed in the reaction and limits the amount of product formed. To find the limiting reactant, we compare the moles of barium and barium oxide to the stoichiometry of the reaction.

Step 3: Calculate the Mass of Barium Hydroxide Formed

Once we have determined the limiting reactant, we can use the stoichiometry of the reaction to calculate the mass of barium hydroxide formed.

I will now calculate the moles of barium and barium oxide and determine the limiting reactant.