I have assigned oxidation numbers to all the elements involved and you can see that none are changed, so this is not a redox reaction. $stackrelcolor(blue)(+2)("Ca")stackrelcolor(blue)(+4)("C")stackrelcolor(blue)(-2)("O")_3+stackrelcolor(blue)(+4)("S")stackrelcolor(blue)(-2)"O"_2rarrstackrelcolor(blue)(+2)("Ca")stackrelcolor(blue)(+4)("S")stackrelcolor(blue)(-2)("O")_3+stackrelcolor(blue)(+4)("C")stackrelcolor(blue)(-2)"O"_2$ As $"SO"_2$ forms $"CaSO"_3$...
1 Answers 1 viewsIf one mole of $O_2$ gives you 2 moles of $SO_3$, then ratio between $O_2 : SO_3$ is 1 : 2
1 Answers 1 viewsSynthesis of sulfur trioxide is represented stoichiometrically: $SO_2(g) + 1/2O_2(g) rarr SO_3(g)$ This equation tells us that $64*g$ of sulfur dioxide reacts with $16$ dioxygen gas to give $80*g$ of...
1 Answers 1 viewsInitially, you have $4.80color(red)(cancel(color(black)("g O"_2))) × "1 mol O"_2/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.1500 mol O"_2$ Thus, you have $"0.1500 mol O"_2/"15.0 L" = "0.0100 mol O"_2/"1 L"$ If you...
1 Answers 1 viewsWe start by finding the relationship between pressure and number of moles. Since temperature is constant, we can consider this:(derived from the Ideal Gas Law, $PV=nRT$: $P_1/n_1=P_2/n_2$, so...
1 Answers 1 viewsHow would you calculate K for the following equilibrium when [SO3] = 0.0160 mol/L, [SO2] = 0.00560 mol/L. and [O2] = 0.00210 mol/L? $K = 3.89 × 10^3$ You...
1 Answers 1 viewsThe equilibrium follows the reaction: $SO_2(g) + NO_2(g) rightleftharpoons SO_3(g) + NO(g)$ And $K_"eq"=3.75=([SO_3(g)][NO(g)])/([SO_2(g)][NO_2(g)])$ And if $x*mol*L^-1$ $SO_2$ reacts................. $K_"eq"=3.75=((0.680+x)(0.680+x))/((0.680-x)(0.680-x))$ $K_"eq"=3.75=(x^2+1.360x+0.462)/(x^2-1.360x+0.462)$ And so, $3.75x^2-5.10x+1.73=x^2+1.360x+0.462$ And thus, $2.75x^2-6.46x+1.268=0$ This has roots...
1 Answers 1 viewsTotal oxidation number of atoms in a compound is zero. The more electronegative element exhibits its typical oxidation number, so in this case oxygen is the more electronegative element, so...
1 Answers 1 viewsAnd for oxides, the oxygen state of the oxygen is necessarily $-II$.. And we assign the oxidation state of the heteroatom on this basis.... $SO_2$, we gots $S(+IV)$; $SO_3$, we...
1 Answers 1 views$SO_3(g) rarr SO_2(g) + 1/2O_2(g)$ Is charge balanced? Is mass balanced? If the answer is no, we cannot accept the reaction as a valid representation of reality?
1 Answers 1 views