Question

instruction:
read each problem carefully and balance the chemical equation. identify the number of moles of the desired substances. use the mole ratio to solve and show the given, find, formula, solution, and final answer (in mol).

1. equation: al + o₂ → __al₂o₃

problem: how many moles of aluminum oxide (al₂o₃) are formed when 6.00 moles of oxygen gas (o₂) react with an excess of aluminum?
2.equation: n₂ + h₂ → __nh₃
problem: how many moles of ammonia (nh₃) are produced when 4.50 moles of hydrogen gas (h₂) react with excess nitrogen gas (n₂)?

1. equation: c₃h₈ + o₂ → co₂ + h₂o

problem: if 2.50 moles of propane (c₃h₈) are completely burned, how many moles of carbon dioxide (co₂) are produced?
4.equation: na + cl₂ → __nacl
problem: how many moles of sodium chloride (nacl) are formed when 5.00 moles of chlorine gas (cl₂) react with excess sodium?

Explanation:

1. Balancing the equation and solving for moles of \(Al_2O_3\)

• Balanced Equation: \(4Al + 3O_2

ightarrow2Al_2O_3\)

• Given: \(n(O_2)=6.00\space mol\)
• Find: \(n(Al_2O_3)\)
• Formula: Mole - ratio from the balanced equation \(\frac{n(Al_2O_3)}{n(O_2)}=\frac{2}{3}\), so \(n(Al_2O_3)=\frac{2}{3}n(O_2)\)
• Solution: \(n(Al_2O_3)=\frac{2}{3}\times6.00\space mol = 4.00\space mol\)

2. Balancing the equation and solving for moles of \(NH_3\)

• Balanced Equation: \(N_2 + 3H_2

ightarrow2NH_3\)

• Given: \(n(H_2)=4.50\space mol\)
• Find: \(n(NH_3)\)
• Formula: Mole - ratio from the balanced equation \(\frac{n(NH_3)}{n(H_2)}=\frac{2}{3}\), so \(n(NH_3)=\frac{2}{3}n(H_2)\)
• Solution: \(n(NH_3)=\frac{2}{3}\times4.50\space mol=3.00\space mol\)

3. Balancing the equation and solving for moles of \(CO_2\)

• Balanced Equation: \(C_3H_8+5O_2

ightarrow3CO_2 + 4H_2O\)

• Given: \(n(C_3H_8)=2.50\space mol\)
• Find: \(n(CO_2)\)
• Formula: Mole - ratio from the balanced equation \(\frac{n(CO_2)}{n(C_3H_8)} = 3\), so \(n(CO_2)=3n(C_3H_8)\)
• Solution: \(n(CO_2)=3\times2.50\space mol = 7.50\space mol\)

4. Balancing the equation and solving for moles of \(NaCl\)

• Balanced Equation: \(2Na+Cl_2

ightarrow2NaCl\)

• Given: \(n(Cl_2)=5.00\space mol\)
• Find: \(n(NaCl)\)
• Formula: Mole - ratio from the balanced equation \(\frac{n(NaCl)}{n(Cl_2)}=2\), so \(n(NaCl) = 2n(Cl_2)\)
• Solution: \(n(NaCl)=2\times5.00\space mol=10.0\space mol\)

Answer:

1. \(4.00\space mol\)
2. \(3.00\space mol\)
3. \(7.50\space mol\)
4. \(10.0\space mol\)