How many atoms of each element are present on reactants (LHS) and product (RSH)

1) 2Cu+O2 = 2CuO LHS = I think is: O 1x2=2 Cu 1x2=2 total 4 RHS O 2x1=2 Cu 2x1=2 Total 4
2) 2H2O = 2H2+O2 LHS = I think is: O 2x1=2 H 2x2=4 total 6 RHS O1x2=2 H2x2=4 total 6
3) BaCl2+Na2S = BaS+2NaCl = LHS I think is: Ba 1x1=1 Cl 1x2=2 Na 1x2=2 S 1x1=1 total 6
RHS I think is: Ba 1x1=1 Cl 2x1=2 Na 2x1=2 S1x1=1 total 6
4) 2C3H8O=9O2 = 8H2O+6CO2 = I GOT REALLY CONFUSED AND DOESN'T MATCH MY TOTALS :( PLEASE HELP
5) N2+3H2=2NH3 =
6) 4Fe+3O2=2Fe2O3 =
7) Au2S3+3H2=3H2S+2Au =
8) C4H10O+6O2=5H2O+4CO2 =

Q1, correct
Q2, correct
Q3, correct
Q4, C3H8O has 3 C, 8 H and 1 O. You have 2 molecules of it so you have 6 C, 16 H and 2 O. You also have 9O2 on the LHS which is 18 O. Total for both is 6 C, 16H and 20 O.

For the RHS, CO2 has 1 C and 2 O. You have 6 molecules of CO2 so you have 6 C and 12 O. For H2O, you have 2H and 1 O. You have 8 H2O which means you have 16H and 8 O. For the two, that gives you 6 C, 20 O and 16H

Q5, LHS has 2N and 6H, RHS has 2N and 6H

Q6, LHS has 4Fe (4 * 1) and 6O (3 * 2), RHS has 4Fe (2 * 2) and 6O (2 * 3)

Q7, Au2S3+3H2=3H2S+2Au

On the LHS you have 2Au, 3S and 6H (3 * 2). On the RHS, you have 6H (3 * 2) + 3S (3 * 1) and 2Au

Q8, C4H10O+6O2=5H2O+4CO2

C4H10O has 4C, 10H and 1O
6O2 has 12O

Total of 4C, 10C and 13O

5H2O has 10H and 5O
4CO2 has 4C and 8O

Total of 4C, 10H and 13O (5 + 8)