Questions: What is the coefficient in front of Na2 S when the chemical equation is correctly balanced Al(NO3)3+ Na2 S -> Al2 S3 + NaNO3

Solution Steps

The given chemical reaction is:

\[ \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} + \mathrm{Na}_{2} \mathrm{~S} \rightarrow \mathrm{Al}_{2} \mathrm{~S}_{3} + \mathrm{NaNO}_{3} \]

• Aluminum (Al):

  • Reactants: 1 (from \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\))
  • Products: 2 (from \(\mathrm{Al}_{2} \mathrm{~S}_{3}\))

• Sodium (Na):

  • Reactants: 2 (from \(\mathrm{Na}_{2} \mathrm{~S}\))
  • Products: 1 (from \(\mathrm{NaNO}_{3}\))

• Sulfur (S):

  • Reactants: 1 (from \(\mathrm{Na}_{2} \mathrm{~S}\))
  • Products: 3 (from \(\mathrm{Al}_{2} \mathrm{~S}_{3}\))

• Nitrate (NO\(_3\)):

  • Reactants: 3 (from \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\))
  • Products: 1 (from \(\mathrm{NaNO}_{3}\))

To balance the aluminum atoms, we need 2 moles of \(\mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\):

\[ 2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} + \mathrm{Na}_{2} \mathrm{~S} \rightarrow \mathrm{Al}_{2} \mathrm{~S}_{3} + \mathrm{NaNO}_{3} \]

To balance the sodium atoms, we need 6 moles of \(\mathrm{NaNO}_{3}\):

\[ 2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} + \mathrm{Na}_{2} \mathrm{~S} \rightarrow \mathrm{Al}_{2} \mathrm{~S}_{3} + 6 \mathrm{NaNO}_{3} \]

To balance the sulfur atoms, we need 3 moles of \(\mathrm{Na}_{2} \mathrm{~S}\):

\[ 2 \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3} + 3 \mathrm{Na}_{2} \mathrm{~S} \rightarrow \mathrm{Al}_{2} \mathrm{~S}_{3} + 6 \mathrm{NaNO}_{3} \]

Final Answer