Balance the following equations: 1. Mg(OH)2 + H3AsO4 → Mg3(AsO4)2 + H2O 2. NH4NO3 → N2O + H2O 3. AsCl3 + H2O → H3AsO3 + HCl 4. C4H10O2 + O2 → CO2 + H2O Use the equation for questions 5- 8: Cl2 + F2 → ClF3, 5. How many moles of Cl2 are needed to react with 3.44 moles of F2? 6. How many grams of ClF3 form when 0.204 moles of F2 react with excess Cl2? 7. How many grams of ClF3 form from 130.0 grams of Cl2 when F2 is in excess?

A. The balanced chemical equations of the given reactions are shown below:

1. 3Mg(OH)₂ + 2H₃AsO₄ → Mg₃(AsO₄)₂ + 6H₂O

2. NH₄NO₃ → N₂O + 2H₂O

3. AsCl₃ + 3H₂O → H₃AsO₃ + 3HCl

4. 2C₄H₁₀O₂ + 11O₂ → 8CO₂ + 10H₂O

B. Number of moles of Cl₂ needed to react with 3.44 moles of F₂ = 1.14 moles

C. mass of ClF₃ produced = 0.136 * 92.5 = 12.58 g

D. 130.0 g of Cl₂ will produce (130 * 92.5)/71 g of ClF₃ = 169.37 g of ClF₃

Explanation:

A balanced chemical equation occurs when the number of the atoms involved in the reactants side is equal to the number of atoms in the products side.This in accordance with the law of conservation of mass which explains that when a chemical reaction occurs, the mass of the products should be equal to the mass of the reactants.

The balanced chemical equations of the given reactions are shown below:

1. 3Mg(OH)₂ + 2H₃AsO₄ → Mg₃(AsO₄)₂ + 6H₂O

2. NH₄NO₃ → N₂O + 2H₂O

3. AsCl₃ + 3H₂O → H₃AsO₃ + 3HCl

4. 2C₄H₁₀O₂ + 11O₂ → 8CO₂ + 10H₂O

B. The balanced equation of the reaction is: Cl₂ + 3F₂  → 2ClF₃

From the equation above, I mole of Cl₂ reacts with 3 moles of F₂

Number of moles of Cl₂ needed to react with 3.44 moles of F₂ will be = 1 * 3.44/3 = 1.14 moles

C. From the equation above,  3 moles of F₂ reacts to produce 2 moles of ClF₃

0.204 moles of F₂ will produce (2 * 0.204)/3 of ClF₃ = 0.136 moles of ClF₃

Molar mass of ClF₃ = 35.5 + 3 * 19 = 92.5 g/mol

mass of ClF₃ produced = number of moles * molar mass

mass of ClF₃ produced = 0.136 * 92.5 = 12.58 g

D. From the equation of reaction, 1 mole of Cl₂ produces 2 moles of ClF₃

molar mass of Cl₂ = 71 g/mol

71 g of Cl₂ produces 2 * 92.5 g of ClF₃

130.0 g of Cl₂ will produce (130 * 92.5)/71 g of ClF₃ = 169.37 g of ClF₃

increased stability in the completion of fluorine's outer shell.

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