Answer to Question #341136 in General Chemistry for abood

Question

A sample of pure diamond was burned in pure oxygen, formed carbon dioxide was collected, and then bubbled through 3.50 L of 0.437 M sodium hydroxide solution. Bubbled carbon dioxide

was converted to sodium carbonate by the reaction shown below.

CO2(a) + NaOH,da) > Na) COgag) + H200)

After the reaction remaining, NaH was determined by back titration with 0.350 M HI solution,

and 1.71 L of the titrant was consumed for neutralization of the leftover NaH. What volume of

oxygen gas was consumed in the process if it was measured at 7.6 atm and 20 °C?

Answer 1

$C+O_2\rightarrow CO_2$

$CO_2 +2 NaOH \rightarrow Na_2CO_3 + H_2O$

$NaOH + HI \rightarrow NaI + H_2O$

$n(HI)=0.35x1.71=0.5985 mole$

$n(NaOH_{leftover})=n (HI)=0.5985 mole$

$n(NaOH_{all})=0.437x3.5=1.5295 mole$

For reaction with CO₂ was used 1.5295-0.5985=0.931 mole of NaOH

$n(CO_2)=n(NaOH_{react}/2)=0.931)2=0.4655 mole$

PV=nRT

P=7.6 atm=770070 Pa

T=20 °C=293 K

R=8.31 J/(K mole)

$V=nRT/P=0.4655(8.31)293/770070=0.00147 m^3$