The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
Answer:
Explanation:
Given
The data in the table
Required
Follow the steps appended to the question;
Step 1: Calculate the Mean or Average
Mean = Summation of lengths divided by number of teams;
Step 2: Get The Range
Step 3: Divide Range by 2
Step 4: Determine the Precision
Substitute 2.70 for Average and 0.1 for Approximate Range
Start with the 19.7 mol HNO3. use dimensional analysis to correctly convert from mol HNO3 to gram H2O. so, it should look similar to 19.7 mol HNO3 x (2 mol H2O/6 mol HNO3) x (18 g H2O/1 mol H2O)
the first parenthesis’ numbers were received from the balanced equation (for every 6 mol HNO3, 2 mol H2O formed). the second is converting from moles to grams by using the molar mass of H2O (1+1+16). you should get 709.2/6. once you divide those, the answer should be 118.2 g H2O. I’m not sure if your computer requires you to use the exact answer or stop at the correct number of significant digits, but if it does then it might just be 118. g H2O.
