which of these is the correct lewis structure for ethyne (acetylene), c₂h₂? h—c(=c)—h (with lone pairs)…

which of these is the correct lewis structure for ethyne (acetylene), c₂h₂? h—c(=c)—h (with lone pairs), :h—c=c—:h (with lone pairs), :c—h—h—:c (with lone pairs), h—c≡c—h

which of these is the correct lewis structure for ethyne (acetylene), c₂h₂? h—c(=c)—h (with lone pairs), :h—c=c—:h (with lone pairs), :c—h—h—:c (with lone pairs), h—c≡c—h

Answer

Problem 3: Balancing ( \boldsymbol{\ce{C3H8 + O2 -> CO2 + H2O}} )

Explanation:

Step 1: Balance Carbon Atoms

There are 3 C in ( \ce{C3H8} ), so put 3 in front of ( \ce{CO2} ):
( \ce{C3H8 + O2 -> 3CO2 + H2O} )

Step 2: Balance Hydrogen Atoms

There are 8 H in ( \ce{C3H8} ), so H in ( \ce{H2O} ) should be 8. Since each ( \ce{H2O} ) has 2 H, put 4 in front of ( \ce{H2O} ):
( \ce{C3H8 + O2 -> 3CO2 + 4H2O} )

Step 3: Balance Oxygen Atoms

Right side: ( 3\times2 + 4\times1 = 10 ) O. So put 5 in front of ( \ce{O2} ) (since ( \ce{O2} ) has 2 O per molecule):
( \ce{C3H8 + 5O2 -> 3CO2 + 4H2O} )

Answer (Problem 3):

( \boldsymbol{\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}} )

Problem 4: Balancing ( \boldsymbol{\ce{C6H10 + O2 -> CO2 + H2O}} )

Explanation:

Step 1: Balance Carbon Atoms

6 C in ( \ce{C6H10} ), so put 6 in front of ( \ce{CO2} ):
( \ce{C6H10 + O2 -> 6CO2 + H2O} )

Step 2: Balance Hydrogen Atoms

10 H in ( \ce{C6H10} ), so H in ( \ce{H2O} ) is 10. Put 5 in front of ( \ce{H2O} ) (2 H per ( \ce{H2O} )):
( \ce{C6H10 + O2 -> 6CO2 + 5H2O} )

Step 3: Balance Oxygen Atoms

Right side: ( 6\times2 + 5\times1 = 17 ) O. Since ( \ce{O2} ) has 2 O, we need ( \frac{17}{2} ), but multiply all coefficients by 2 to eliminate fractions:
( \ce{2C6H10 + 17O2 -> 12CO2 + 10H2O} )

Answer (Problem 4):

( \boldsymbol{\ce{2C6H10 + 17O2 -> 12CO2 + 10H2O}} )

Problem 5: Balancing ( \boldsymbol{\ce{Al + CuCl2 -> AlCl3 + Cu}} )

Explanation:

Step 1: Balance Chlorine Atoms

2 Cl in ( \ce{CuCl2} ), 3 Cl in ( \ce{AlCl3} ). LCM of 2 and 3 is 6. So put 3 in front of ( \ce{CuCl2} ) and 2 in front of ( \ce{AlCl3} ):
( \ce{Al + 3CuCl2 -> 2AlCl3 + Cu} )

Step 2: Balance Aluminum Atoms

2 Al on right, so put 2 in front of ( \ce{Al} ):
( \ce{2Al + 3CuCl2 -> 2AlCl3 + Cu} )

Step 3: Balance Copper Atoms

3 Cu on left, so put 3 in front of ( \ce{Cu} ):
( \ce{2Al + 3CuCl2 -> 2AlCl3 + 3Cu} )

Answer (Problem 5):

( \boldsymbol{\ce{2Al + 3CuCl2 -> 2AlCl3 + 3Cu}} )

Problem 6: Balancing ( \boldsymbol{\ce{Na3PO4 + CaCl2 -> Ca3(PO4)2 + NaCl}} )

Explanation:

Step 1: Balance Phosphate (( \boldsymbol{\ce{PO4^{3-}}} )) Ions

2 ( \ce{PO4^{3-}} ) in ( \ce{Ca3(PO4)2} ), so put 2 in front of ( \ce{Na3PO4} ):
( \ce{2Na3PO4 + CaCl2 -> Ca3(PO4)2 + NaCl} )

Step 2: Balance Calcium Atoms

3 Ca in ( \ce{Ca3(PO4)2} ), so put 3 in front of ( \ce{CaCl2} ):
( \ce{2Na3PO4 + 3CaCl2 -> Ca3(PO4)2 + NaCl} )

Step 3: Balance Sodium and Chloride Atoms

Left: 6 Na (from ( 2\times3 )) and 6 Cl (from ( 3\times2 )). So put 6 in front of ( \ce{NaCl} ):
( \ce{2Na3PO4 + 3CaCl2 -> Ca3(PO4)2 + 6NaCl} )

Answer (Problem 6):

( \boldsymbol{\ce{2Na3PO4 + 3CaCl2 -> Ca3(PO4)2 + 6NaCl}} )

Note on Problem 2 (Completed in Image):

The balanced equation for ( \ce{S8 + F2 -> SF6} ) is ( \boldsymbol{\ce{S8 + 24F2 -> 8SF6}} ) (matches the image’s work: 8 S on left/right, ( 24\times2 = 48 ) F on left, ( 8\times6 = 48 ) F on right).