The type of problems involved here can be solved easily with 23 simple steps. Use the Bridge Method for solving only after the equation is balanced. 
Never, ever assume that an equation is balanced.

Example Solution #1 
During its combustion, ethane C_{2}H_{6}, combines with oxygen O_{2} to give carbon dioxide and water. A sample of ethane was burned completely and the water that formed has a mass of 1.61 grams. How much ethane, in moles and in grams, was in the sample? 
Solution: 
1. Set up the equation based on the words in the problem.
Then balance it correctly.
2 C_{2}H_{6} + 7 O_{2}
> 4 CO_{2} + 6 H_{2}O

2. Compute the needed formula molecular masses.
We need the molecular mass for water and for
ethane. H_{2}O = 18.02 g.mol; C_{2}H_{6}
= 30.08 g/mol 
3. Draw a roadmap outline the math steps that need to
be taken:
moles of water > moles of ethane 
Therefore this type of question requires a three step solution.
Step #1 Convert 1.61 grams of water to moles of water. Step #2 Using the equation, compare the moles of water made from moles of ethane. Step #3 Convert the moles of ethane back into grams of ethane. 
Draw a roadmap outlining the mathematical steps that need to be
taken:
moles of water > moles of ethane 
Math Step #1 Convert 1.61 grams of water into moles of water 
Moles = g/mm = 1.61 g/ 18.02 g/mol = 0.09 mol of water were used. 
Math Step #2 The chemical equation shows us that 2 moles of C_{2}H_{6} is needed to make 6 moles of water. Setup a ratio: 
6 mol H_{2}O = 2 mol C_{2}H_{6} 0.09 mol x x = 0.03 moles of ethane are needed. 
Math Step #3 Convert the moles of ethane back into grams of ethane. 
g = n * mm = 0.03 mol * 30.08 g/mol = 0.90 grams of ethane 
Finish off the question with a statement. 1.61 grams of water can be made from 0.03 moles of ethane or 0.90 grams of ethane. 
Example Question #2 
Calculate how many grams of K_{2}Cr_{2}O_{7} are needed to make 35.8 grams of I_{2} according to the following equation. 
K_{2}Cr_{2}O_{7} + 6 NaI +
7 H_{2}SO_{4} > Cr_{2}(SO_{4})_{3} + 3 I_{2} + 7 H_{2}O + 3 Na_{2}SO_{4} + K_{2}SO_{4} 
The equation is already balanced but you should check it over just to be sure. 
Next calculate the molecular masses that you need. The question
deals with iodine, I_{2}, and potassium dichromate, K_{2}Cr_{2}O_{7}.
Calculate the molecular masses for these only. The other parts of
the equation can be forgotten about.
I_{2} = 253.82 g/mol K_{2}Cr_{2}O_{7} = 294.20 g/mol 
Draw a roadmap outlining the mathematical steps that need to be
taken:
moles of iodine > moles of potassium dichromate

Math Step #1 Convert grams of iodine into moles of iodine.
moles = g/mm = 35.8 g / 253.82 g/mol = 0.14 moles of iodine 
Math Step #2 Compare using the equation coefficients, the
number of moles of potassium dichromate is need to make how many moles of
iodine.
3 I_{2} = 1 K_{2}Cr_{2}O_{7}
x = 0.05 moles of potassium dichromate 
Math Step #3 Convert the moles of potassium dichromate back into
grams of potassium dichromate.
g = n * mm = 0.05 mol * 294.20 g/mol = 14.71 grams To make 35.8 grams of iodine you must start with 14.71 grams of
potassium dichromate. 