Question

In the reaction BaCO3 + 2HNO3 + Ba(NO3)2 + CO2 + H2O, what mass of Ba(NO3)2 can be formed by combining 55 g BaCO3 and 26 g HNO3

Answer 1

The mass of Ba(NO3)2 that can be formed from the reaction of 55 g BaCO3 and 26 g HNO3 is calculated to be approximately 53.99 g based on stoichiometry, with HNO3 as the limiting reactant.

Step-by-step explanation:

The question is related to a chemical reaction between barium carbonate and nitric acid, producing barium nitrate, carbon dioxide, and water. To calculate the mass of Ba(NO3)2 formed, we need to perform a stoichiometric calculation based on the balanced chemical reaction. The balanced equation is:

BaCO3(s) + 2HNO3(aq) → Ba(NO3)2(aq) + CO2(g) + H2O(l)

First, we should determine the number of moles of the reactants. The molar mass of BaCO3 is 197.34 g/mol, and the molar mass of HNO3 is 63.01 g/mol. There are 55 g of BaCO3 and 26 g of HNO3:

1. Calculate moles of BaCO3: 55 g ÷ 197.34 g/mol ≈ 0.279 moles of BaCO3
2. Calculate moles of HNO3: 26 g ÷ 63.01 g/mol ≈ 0.413 moles of HNO3

According to the balanced equation, 1 mole of BaCO3 reacts with 2 moles of HNO3 to produce 1 mole of Ba(NO3)2. Because there is a limited amount of HNO3 available, it will be the limiting reactant. Since 1 mole of HNO3 produces 0.5 mole of Ba(NO3)2, we can calculate the mass of Ba(NO3)2 formed:

1. Moles of Ba(NO3)2 produced from HNO3: 0.413 moles × 0.5 ≈ 0.2065 moles of Ba(NO3)2
2. Calculate mass of Ba(NO3)2: 0.2065 moles × 261.34 g/mol (molar mass of Ba(NO3)2) ≈ 53.99 g

Thus, 53.99 g of Ba(NO3)2 can be formed.

Answer 2

Answer: IT IS NOT BA(NO3)2

Step-by-step explanation:

WHAT THE OTHER GUY WROTE WAS TOO CONFUSING POEPLE NEED TO BUT THE ANSWE IN A CLEAR LOCATION. I FAILED THIS TEST