Question

Are these equations balanced correctly? Zn(s) + 2

Na+(aq) + 2 OH-(aq) + 2 H2O(l) → Zn(OH)42-(aq) + 4 Na+(aq) + H2(g)

2 C3H6(g) + 5 O2(g) → 6 CO2(g) + 8 H2O(l)

Answer 1

2nd one no. YOu have 12 H on left ond 16 on right. You have 10 O on left and 20 on right.

1st one no. You have
2 Na ions on left and 4 Na ions on right.
zero charge on left and 2+ charge on right.

Answer 2

Zn(s) + Na+(aq) + 2 OH-(aq) + 2 H2O(l) → Zn(OH)42-(aq) + Na+(aq) + H2(g)

2C3H6(g) + 4O2(g) → 6CO2(g) + 6H2O(l)

Would this be right?

Answer 3

I don't understand. You can check these yourself and know if you are right or wrong without asking any help.
1. looks ok to me.
2. Count the O atoms. I count 8 O atoms on the left and I count 18 on the right.
Here is how I do them.
C3H6 + O2 ==> CO2 + H2O

I need 3C on the right so I place a coefficient of 3 for CO2 on the right.
C3H6 + O2 --> 3CO2 + H2O

How many H do I need on the right? 6 so I place a 3 in front of H2O like this.
C3H6 + O2 --> 3CO2 + 3H2O

C is ok. H is ok. Now I count up O on the right which I need (because I can make that any number I wish on the left). I need 6 O from CO2 and 3 from H2O for a total of 9 O atoms. So I multiply O2 by 9/2 (9/2 x 2 = 9).
C3H6 + 9/2 O2 ==> 3CO2 + 3H2O

We usually don't like to keep fractions in the final equation so I multiply the entire equation through by 2.
2C3H6 + 9 O2 ==> 6CO2 + 6H2O
Now we check it.
6C on left and 6 on right.
12 H on left and 12 on right.
18 O on left and 18 on right (6*2=12 from 6CO2 and 6*1=6 from 6H2O and 12+6=18 total).