Using the following equation for the combustion of octane, calculate the amount of grams of carbon dioxide formed from 100.0 g of octane. the molar mass of octane is 114.33 g /mole .the molar mass of carbon dioxide is 44.0095 g/mole.
2 c8h18 +25 o2 > 16 co2 +18 h20
/\h 11018 kj

307.94g of CO₂

Explanation:

Equation of reaction

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

molarmass of C₈H₁₈ = 114.33g/mol

2 moles of C₈H₁₈ = 228.66g/mol

Molar mass of CO₂ = 44.0095g/mol

16 moles of CO₂ = 704.152g/mol

From equation of reaction,

2 moles of C₈H₁₈ reacts with 16 moles of CO₂;

228.66g of C₈H₁₈ = 704.152g of CO₂

100g of C₈H₁₈ = y moles of CO₂

y = (100 * 704.152) / 228.66

y = 307.94g of CO₂

Therefore, 100g of C₈H₁₈would yield 307.94g of CO₂.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"