Calculate the mass of water produced when 3.09g of butane reacts with excess oxygen.

Question

2C4H10 + 13O2-->10H2O + 8CO2

Help.

Answer 1

You need to copy that format I gave you earlier. All of these stoichiometry problems are worked the same way.
2C4H10 + 13O2-->10H2O + 8CO2

Step 1. Write and balance the equation. You have that.

Step 2. Convert what you have (in this case butane) into mols. mols = grams/molar mass = 3.09/5 = estd 0.05

Step 3. Using the coefficients in the balanced equation convert mols of what you have (butane) to mols of what you want (in this case H2O)
0.05 mols butane x (10 mols H2O/2 mols butane) = 0.05 x 10/2 = estd 0.27

Step 4. Now convert that to grams or whatever unit you wish.
grams = mols x molar mass or
g = 0.27 x 18 = estd 4.79 grams C4H10

Answer 2

You need to copy that format I gave you earlier. All of these stoichiometry problems are worked the same way.
2C4H10 + 13O2-->10H2O + 8CO2

Step 1. Write and balance the equation. You have that.

Step 2. Convert what you have (in this case butane) into mols. mols = grams/molar mass = 3.09/58 = estd 0.05

Step 3. Using the coefficients in the balanced equation convert mols of what you have (butane) to mols of what you want (in this case H2O)
0.05 mols butane x (10 mols H2O/2 mols butane) = 0.05 x 10/2 = estd 0.27

Step 4. Now convert that to grams or whatever unit you wish.
grams = mols x molar mass or
g = 0.27 x 18 = estd 4.79 grams C4H10