2C4H10 + 13O2 = 8CO2 + 10H2O
1. (2.06g C4H10)/(58.12 g/mol C4H10) = 0.035mol C4H10
2. (0.035molC4H10)(10 mol H2O/2mol C4H10) = 0.177mol H2O
3. (0.177mol H2O)(18.01g/mol H2O) = 3.19g H2O
Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
<span><span>The reaction is as follows:
C6H6 </span>+ HNO3 + H2SO4 ------------> </span>C6H5NO2<span> + H</span>2<span>O
(BENZENE) (NITRIC ACID)(CATALYST)
</span>NO2(+) is the electrophile that acctacks on the benzene ring in nitration process.
Answer:
1 kiloliter = 1,000,000 milliliters
45 × 1,000,000
45 KL = 45,000,000 ML
