Question

question 21 of 37
balance the following chemical equation (if necessary):
v_{2}o_{5}(s) + 10 hcl(aq) \to 2 vocl_{3}(s) + 5 h_{2}o(l)

Explanation:

Step1: Count the number of each atom on both sides

• Vanadium (\(V\)):
• Left side: \(2\) (from \(V_2O_5\))
• Right side: \(2\) (from \(2VOCl_3\))
• Oxygen (\(O\)):
• Left side: \(5\) (from \(V_2O_5\))
• Right side: \(2\times1 + 5\times1=7\) (from \(2VOCl_3\) and \(5H_2O\))
• Hydrogen (\(H\)):
• Left side: \(10\) (from \(10HCl\))
• Right side: \(5\times2 = 10\) (from \(5H_2O\))
• Chlorine (\(Cl\)):
• Left side: \(10\) (from \(10HCl\))
• Right side: \(2\times3=6\) (from \(2VOCl_3\))

Step2: Adjust the coefficients

• For oxygen balance:
• We have \(V_2O_5\) on the left. Let's check the oxygen - hydrogen - chlorine relationships.
• The original equation \(V_2O_5(s)+10HCl(aq)\to2VOCl_3(s)+5H_2O(l)\)
• The correct balanced equation is \(V_2O_5(s)+6HCl(aq)\to2VOCl_3(s)+3H_2O(l)\)
• Vanadium (\(V\)):
• Left side: \(2\) (from \(V_2O_5\))
• Right side: \(2\) (from \(2VOCl_3\))
• Oxygen (\(O\)):
• Left side: \(5\) (from \(V_2O_5\))
• Right side: \(2\times1+3\times1 = 5\) (from \(2VOCl_3\) and \(3H_2O\))
• Hydrogen (\(H\)):
• Left side: \(6\) (from \(6HCl\))
• Right side: \(3\times2=6\) (from \(3H_2O\))
• Chlorine (\(Cl\)):
• Left side: \(6\) (from \(6HCl\))
• Right side: \(2\times3 = 6\) (from \(2VOCl_3\))

Answer:

\(V_2O_5(s)+6HCl(aq)\to2VOCl_3(s)+3H_2O(l)\)