How do you find the percent composition of all elements in sulfur trioxide?

We need a stoichiometic equation: $SO_2(g) + 1/2O_2(g) rarr SO_3(g)$ The equation unequivocally tells us that the reaction of $64*g$ $SO_2(g)$ with $16*g$ $O_2(g)$ gives $80*g$ $SO_3(g)$. The given masses...

Sulfur assumes a $+VI$ oxidation state in the acid anhydride sulfur trioxide, and of course has the same oxidation state in sulfuric acid. $(O=)S^(2+)(-O^-)_2$ Most chemists would settle for...

.........And oxygen is in deficiency. Given $10$ $mol$ of dioxygen gas, at most, $6.67$ $mol$ of sulfur trioxide can be generated. This is clearly indicated by the stoichometric equation. Should...

Vast quantities of $"sulfur trioxide"$ are utilized industrially for sulfuric acid synthesis: $SO_3 + H_2Orightleftharpoons H_2SO_4$

We can combine the given equations........ $S(s) + 3/2O_2(g) rarr SO_3(g)$ And this clearly gives us the mass transfer; 1 equiv of sulfur is oxidized by 3/2 equiv dioxygen gas........

$"Moles of sulfur"$ $=$ $(10.0*g)/(32.06*g*mol^-1)=0.312*mol$. Given the , we require $3/2xx0.312*mol$ of dioxygen gas, i.e. $0.468*mol$. And now we solve for volume in the Ideal Gas equation: $V=(nRT)/P$ $=$ $(0.468*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx623*cancel(K))/(5.25*cancel(atm))$...

$"8.5 mols of gas"$ Write the reaction: $2"SO"_2(g) + "O"_2(g) -> 2"SO"_3(g)$ For this reaction, you know that the starting mols are: $n_("SO"_2,i) = "5 mols...

$4As$ + $ 3O_2$ -----> $2As_2O_3$ According to equation $3$ moles of oxygen gas is required to form arsenic trioxide = $2$ moles $3.4$ moles of O2 will be required...

We thus predict that hydrogen sulfide would have a formula of $H_2S$, an analogue of the popular $H_2O$ molecule.

These are just conversions. Set up the equation so that you end up with the UNIT you desire for an answer, and the math will be correct. $(1/32.07)$ (atom/amu) *...