Question

I Review I Constants I Periodic Table When a chemical reaction occurs, atoms rearrange to form new compounds, but no new atoms are created nor are any destroyed. This concept is called conservation of Part A mass. Mass conservation can be seen in a balanced chemical equation, where the numbers of each kind of atom are the same on both sides of the reaction arrow. The figure (Figure 1) shows the reaction of element A (lavender spheres) with element B (tan spheres). Write the balanced chemical equation for this reaction in terms of A and B Express your answer as a chemical equation. View Available Hint(s) Figure 1 of 1 ΑΣφ Submit Balancing equations with polyatomic ions When balancing chemical equations with polyatomic ions, it may help to balance the ion as a whole. For example, the ion PO4can be treated as a unit and balanced in both the reactants and the product, or you can break it down and balance each atom separately. Part B

Please answer al three questions. Thank you!!

Answer 1

PART A

From given figure it is clear that , 4 molecules of A combines with 6 molecules of B 2 to form 4  molecules of AB ₃ . Hence, we can write equation as shown below.

4 A + 6 B ₂$\rightarrow$ 4 AB ₃

We can write above equation by adjusting coefficients to lowest value.

2 A + 3 B ₂$\rightarrow$ 2 AB ₃

ANSWER : 2 A + 3 B ₂$\rightarrow$ 2 AB ₃

PART B

Consider a given equation, H₃PO ₄ + Mg(OH) ₂ $\rightarrow$ Mg₃(PO ₄) ₂ + H₂O

Mg₃(PO ₄) ₂ contain three Mg atoms and two phosphate groups, hence to balance equation adjust coefficient of magnesium hydroxide to 3 and that of phosphoric acid to 2.

2 H₃PO ₄ + 3 Mg(OH) ₂ $\rightarrow$ Mg₃(PO ₄) ₂ + H₂O  

Now balance hydrogen atoms. There are 12 hydrogen are present L.H.S of equation & 2 on R.H.S of equation.

To balance Hydrogen atoms adjust coefficient of water to 6.

2 H₃PO ₄ + 3 Mg(OH) ₂ $\rightarrow$ Mg₃(PO ₄) ₂ + 6 H₂O  

Above equation is balanced for all atoms.

PART C

CaCO ₃ (s) + 2 HCl (aq) $\rightarrow$ CaCl ₂ (aq) + H₂O (l) + CO ₂ (g)  

Given : Mass of CaCO ₃ = 20.00 g

Mass of HCl solution = 81.00 g

$\therefore$ Total mass of reactants = 20.00 + 81.00 = 101.00 g

From the law of conservation of mass, we can write Total mass of reactants = Total mass of products.

$\therefore$ 101.00 g = Total mass of products.

We have, Total mass of products = Mass of CaCl ₂ + Mass of  H₂O (l) + Mass of CO ₂ (g)  

We have given information that, (Mass of CaCl ₂ + Mass of  H₂O (l)) = 92.80 g

$\therefore$ 101.00 g = 92.80 g + Mass of CO ₂ (g)  

Mass of CO ₂ (g) = 101.00 g - 92.80 g = 8.20 g

ANSWER : Mass of CO ₂ produced in the reaction = 8.20 g