## Test 1 Q3

Bree Perkins 1E
Posts: 34
Joined: Fri Apr 06, 2018 11:04 am

### Test 1 Q3

Hi, I was wondering if someone could help me balance and equation. The hard part for me to understand is when you have to use a fraction and the multiply to get whole numbers. The equation is below:

$Cu + HNO_{3} \rightarrow Cu(NO_{3})_{2} + NO + H_{2}O$

NabilaNizam-1K
Posts: 30
Joined: Fri Apr 06, 2018 11:04 am

### Re: Test 1 Q3

Answer: $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 2NO +4H_2O$

Usually, for balancing equation questions, I try balancing the least abundant element first which is Cu in this case. Hope this helps!

Alejandro Salazar 1D
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: Test 1 Q3

It is hard to know when to use fractions especially on this one.

But if you consider a more simpler equation like C2H6 +02 --> CO2+H20

C2H6+O2 --> 2CO2+ 3H2O

C:2 C:2
H:6 H:6
O:2 O:7

There is no number you can multiply two from the oxygen on the reactant side to get 7 from the product side. I find it easier if you think of it like this. 2x=7 . Solve for x and you get 7/2 (your coefficient fraction)

C2H6+ 7/2(O2) ---> 2CO2+3H2O

Now multiply by 2 to get rid of the denominator.

2H2H6 +7O2 --> 4CO2+6H2O