The empirical formula of benzene is ch (its molecular formula os c6h6). if 10.00 mg of benzene is subjected to combustion analysis, what mass of co2 and h2o would be

There would be produced 0.00691 g of water and 0.034 g of CO₂

Explanation:

Combustion for benzene is:

2C₆H₆  +  15O₂  →  12CO₂  +  6H₂O

2 moles of benzene react with 15 moles of oxygen to produce 6 moles of water and 12 moles of carbon dioxide.

Let's convert the mg to moles; previously we should convert mg to g

10 mg = 0.010 g

0.01 g / 78 g/m = 1.28×10⁻⁴ moles

Let's identify ratios. Ratio with CO₂ is 2:12 and 2:6 with water

2 moles of benzene react to produce 12 moles of CO₂

Then 1.28×10⁻⁴ mol of C₆H₆ would produce (1.28×10⁻⁴ . 12)/ 2 = 7.69×10⁻⁴ moles of CO₂.

2 moles of benzene react to produce 6 moles of water

Then 1.28×10⁻⁴ mol of C₆H₆ would produce (1.28×10⁻⁴ . 6) /2 = 3.84×10⁻⁴  moles of water

Let's convert the moles to mass (mol . molar mass)

7.69×10⁻⁴ moles of CO₂  . 44g/m = 0.034 g

3.84×10⁻⁴  moles of water . 18 g/m = 0.00691 g

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol