Question

Обчисліть масу барій оксиду, яка необхідна для одержання 7,88 г BaCO3 внаслідок реакції барій оксиду з карбон (IV) оксидом​

Answer 1

Ответ:

6.12г

Объяснение:

Answer 2

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Your message seems to be in Ukrainian. It translates to: "Answer in detail. Calculate the mass of barium oxide required to obtain 7.88 g of BaCO3 as a result of the reaction of barium oxide with carbon (IV) oxide."

To answer your question, I will use the following balanced chemical equation:

$$BaO + CO_2 \rightarrow BaCO_3$$

According to this equation, one mole of BaO reacts with one mole of CO2 to produce one mole of BaCO3. Therefore, the mole ratio of BaO to BaCO3 is 1:1.

To calculate the mass of BaO needed, we need to convert the given mass of BaCO3 to moles, then use the mole ratio to find the moles of BaO, and then convert the moles of BaO to mass.

The molar mass of BaCO3 is 197.34 g/mol. The molar mass of BaO is 153.33 g/mol.

The moles of BaCO3 are:

$$\frac{7.88 \text{ g}}{197.34 \text{ g/mol}} = 0.0399 \text{ mol}$$

The moles of BaO are:

$$0.0399 \text{ mol} \times \frac{1 \text{ mol BaO}}{1 \text{ mol BaCO3}} = 0.0399 \text{ mol}$$

The mass of BaO are:

$$0.0399 \text{ mol} \times 153.33 \text{ g/mol} = 6.12 \text{ g}$$

Therefore, the mass of barium oxide required to obtain 7.88 g of BaCO3 is 6.12 g.

I hope this answer helps you. If you want to learn more about stoichiometry, you can check out these web pages: [8.6: Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Masses of Reactants](https://chem.libretexts.org/Courses/can/intro/08%3A_Quantities_in_Chemical_Reactions/8.6%3A_Limiting_Reactant_Theoretical_Yield_and_Percent_Yield_from_Initial_Masses_of_Reactants), [ChemTeam: Stoichiometry: Mass-Mass Examples](https://www.chemteam.info/Stoichiometry/Mass-Mass.html), [How many mL of 0.572 M HI are needed to dissolve 5.45 g of BaCO3](https://socratic.org/questions/how-many-ml-of-0-572-m-hi-are-needed-to-dissolve-5-45-g-of-baco3).