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# Final NB 2016-2017 Turley

## Step 5 Balance the

Step 5 Balance the hydrogen atoms next. You have 8 on the left side, so you'll need 8 on the right side. C3H8 + O2 --> 4H2O + 3CO2 On the right side, we added a 4 as the coefficient because the subscript showed that we already had 2 hydrogen atoms. When you multiply the coefficient 4 times the subscript 2, you end up with 8. Step 6 Finish by balancing the oxygen atoms. Because we've added coefficients to the molecules on the right side of the equation, the number of oxygen atoms has changed. We now have 4 oxygen atoms in the water molecule and 6 oxygen atoms in the carbon dioxide molecule. That makes a total of 10 oxygen atoms. Add a coefficient of 5 to the oxygen molecule on the left side of the equation. You now have 10 oxygen molecules on each side. C3H8 + 5O2 --> 4H2O + 3CO2. The carbon, hydrogen and oxygen atoms are balanced. Your equation is complete. 92

1) ___ 2 NaNO3 + ___ PbO ___ Pb(NO3)2 + ___ Na2O 2x Na 1 N 1 O 3+1 Pb 1 Na 2 N 2 O 7 Pb1 2) ___ 6 AgI + ___ Fe2(CO3)3 ___ 2 FeI3 + ___ 3 Ag2CO3 6x Ag 1 L 1 Fe 2 C 3 O 9 Ag 2 L 3 Fe 1 C 1 O3 2x 3x 3) ___ C2H4O2 + ___ 2 O2 ___ 2 CO2 + ___ 2 H2O 2X C 2 H 4 O 2 O 2 C 1 H 2 O 2 O 1 2X 4) ___ ZnSO4 + ___ Li2CO3 ___ ZnCO3 + ___ Li2SO4 Already Balanced Zn 1 S 1 O 4 Li 2 C 1 O 3 Zn 1 S 1 O 4 Li 2 C 1 O 3 5) ___ V2O5 + ___ 5 CaS ___ 5 CaO + ___ V2S5 5x V 2 V 2 0 1 O 5 5x Ca 1 Ca 1 S 5 S 1 93

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