Equation Balancing Which Reactant Has A Coefficient Of 3

Balancing chemical equations is a fundamental concept in chemistry, ensuring that the number of atoms for each element is the same on both sides of the equation, thus adhering to the law of conservation of mass. When faced with multiple chemical equations, determining which one, upon balancing, will have a coefficient of 3 in front of any of the reactants requires a systematic approach. This involves examining each equation individually, applying balancing techniques, and identifying the resulting coefficients. Let's delve into each equation provided and explore the balancing process to arrive at the solution.

Balancing Chemical Equations: A Step-by-Step Approach

Before we dive into the specific equations, it's crucial to understand the methodology behind balancing chemical equations. The primary goal is to ensure that the number of atoms of each element is identical on both the reactant and product sides. This is achieved by placing coefficients in front of the chemical formulas. These coefficients act as multipliers, indicating the number of moles of each substance involved in the reaction. The most common method is balancing by inspection which involves a step-by-step process:

1. Write the unbalanced equation: This is the starting point, representing the chemical reaction without considering the quantitative relationships.
2. Identify the elements present: List all the elements that appear in the equation. This helps in organizing the balancing process.
3. Start with the most complex molecule: Begin by balancing the element that appears in the most complex molecule first. This often simplifies the process.
4. Adjust coefficients: Use coefficients to balance the number of atoms of each element, working systematically through the equation. This step usually needs patience and multiple attempts.
5. Check for balance: Verify that the number of atoms for each element is the same on both sides of the equation. If not, repeat the adjustment process.
6. Simplify coefficients: If all elements are balanced, ensure that the coefficients are in the simplest whole-number ratio. This may involve dividing all coefficients by their greatest common factor.

Following this method ensures the balanced equation accurately represents the stoichiometry of the chemical reaction.

Analyzing the First Equation: $Zn + HCl

ightarrow ZnCl_2 + H_2$

Let's begin by examining the first equation: $Zn + HCl ightarrow ZnCl_2 + H_2$. This reaction represents the interaction between zinc metal (Zn) and hydrochloric acid (HCl), resulting in the formation of zinc chloride ($ZnCl_2$) and hydrogen gas ($H_2$).

1. Unbalanced Equation: $Zn + HCl ightarrow ZnCl_2 + H_2$
2. Elements Present: Zinc (Zn), Hydrogen (H), and Chlorine (Cl).
3. Balancing:
   • Zinc is already balanced with one atom on each side.
   • Chlorine is unbalanced, with one atom on the reactant side (HCl) and two on the product side ($ZnCl_2$). To balance chlorine, we place a coefficient of 2 in front of HCl: $Zn + 2HCl ightarrow ZnCl_2 + H_2$.
   • Now, hydrogen is also balanced with two atoms on each side.
4. Balanced Equation: $Zn + 2HCl ightarrow ZnCl_2 + H_2$

In the balanced equation, the coefficients are 1 for Zn, 2 for HCl, 1 for $ZnCl_2$, and 1 for $H_2$. None of the reactants have a coefficient of 3. Therefore, this equation does not meet the specified criterion.

Analyzing the Second Equation: $H_2SO_4 + B(OH)_3

ightarrow B_2(SO_4)_3 + H_2O$

Next, we consider the equation: $H_2SO_4 + B(OH)_3 ightarrow B_2(SO_4)_3 + H_2O$. This equation represents the reaction between sulfuric acid ($H_2SO_4$) and boric acid ($B(OH)_3$), producing boron sulfate ($B_2(SO_4)_3$) and water ($H_2O$). This equation is more complex and requires a more careful balancing approach.

1. Unbalanced Equation: $H_2SO_4 + B(OH)_3 ightarrow B_2(SO_4)_3 + H_2O$
2. Elements Present: Hydrogen (H), Sulfur (S), Oxygen (O), and Boron (B).
3. Balancing:
   • Start by balancing boron. There is one boron atom in $B(OH)_3$ and two boron atoms in $B_2(SO_4)_3$. Place a coefficient of 2 in front of $B(OH)_3$: $H_2SO_4 + 2B(OH)_3 ightarrow B_2(SO_4)_3 + H_2O$.
   • Next, balance the sulfate ($SO_4$) groups. There is one sulfate group on the reactant side and three on the product side. Place a coefficient of 3 in front of $H_2SO_4$: $3H_2SO_4 + 2B(OH)_3 ightarrow B_2(SO_4)_3 + H_2O$.
   • Now, balance the hydrogen atoms. There are 3 x 2 = 6 hydrogen atoms from sulfuric acid and 2 x 3 = 6 hydrogen atoms from boric acid, totaling 12 hydrogen atoms on the reactant side. On the product side, hydrogen is only present in water. To balance, place a coefficient of 6 in front of $H_2O$: $3H_2SO_4 + 2B(OH)_3 ightarrow B_2(SO_4)_3 + 6H_2O$.
   • Finally, check if oxygen is balanced. On the reactant side, there are 3 x 4 = 12 oxygen atoms from sulfuric acid and 2 x 3 = 6 oxygen atoms from boric acid, totaling 18 oxygen atoms. On the product side, there are 3 x 4 = 12 oxygen atoms from boron sulfate and 6 oxygen atoms from water, also totaling 18 oxygen atoms. Oxygen is balanced.
4. Balanced Equation: $3H_2SO_4 + 2B(OH)_3 ightarrow B_2(SO_4)_3 + 6H_2O$

In the balanced equation, the coefficient in front of sulfuric acid ($H_2SO_4$) is 3. This equation satisfies the specified criterion.

Analyzing the Third Equation: $Fe + AgNO_3

ightarrow Fe(NO_3)_3$

Lastly, let's analyze the equation: $Fe + AgNO_3 ightarrow Fe(NO_3)_3$. This represents the reaction between iron (Fe) and silver nitrate ($AgNO_3$), resulting in the formation of iron(III) nitrate ($Fe(NO_3)_3$) and silver (Ag). This equation involves nitrate groups, which can be balanced as a single unit.

1. Unbalanced Equation: $Fe + AgNO_3 ightarrow Fe(NO_3)_3 + Ag$
2. Elements Present: Iron (Fe), Silver (Ag), Nitrogen (N), and Oxygen (O).
3. Balancing:
   • Start by balancing the nitrate ($NO_3$) groups. There is one nitrate group on the reactant side and three nitrate groups on the product side. Place a coefficient of 3 in front of $AgNO_3$: $Fe + 3AgNO_3 ightarrow Fe(NO_3)_3 + Ag$.
   • Next, balance silver. There are three silver atoms on the reactant side and one on the product side. Place a coefficient of 3 in front of Ag: $Fe + 3AgNO_3 ightarrow Fe(NO_3)_3 + 3Ag$.
   • Iron is already balanced with one atom on each side.
4. Balanced Equation: $Fe + 3AgNO_3 ightarrow Fe(NO_3)_3 + 3Ag$

In the balanced equation, the coefficient in front of silver nitrate ($AgNO_3$) is 3. This equation also satisfies the specified criteria. However, the questions want the coefficient 3 in front of any of the reactants, and in the previous equation, the first reactant $H_2SO_4$ already satisfies the condition.

Conclusion

Upon balancing the given chemical equations, we find that the equation $3H_2SO_4 + 2B(OH)_3 ightarrow B_2(SO_4)_3 + 6H_2O$ has a coefficient of 3 in front of one of the reactants, specifically sulfuric acid ($H_2SO_4$). The equation $Fe + 3AgNO_3 ightarrow Fe(NO_3)_3 + 3Ag$ also has a coefficient of 3 in front of a reactant ($AgNO_3$), but the second equation is the first equation to satisfy the condition. Therefore, the balanced equation with a coefficient of 3 in front of a reactant is: $3H_2SO_4 + 2B(OH)_3 ightarrow B_2(SO_4)_3 + 6H_2O$. Understanding the systematic approach to balancing chemical equations is crucial in determining the stoichiometric relationships between reactants and products, which is a cornerstone of chemical calculations and problem-solving.