According to the following reaction, how many moles of sulfur trioxide will be formed upon the complete reaction of 0.971 moles of sulfur dioxide with excess oxygen gas? sulfer dioxide (g) + oxygen (g) $->$ sulfer trioxide

Thiosulfate, $S_2O_3^(2-)$ has formal $stackrel(-II)S$ and $stackrel(+VI)S$ oxidation states, i.e. an average oxidation state of $(-II+VI)/2=+II$. It is reduced to $stackrel(0)S$, i.e. zerovalent sulfur, and oxidized to $stackrel(+IV)"SO"_2$. And thus...

.........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...

$4As$ + $ 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...

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

From the equation above, 3 moles of carbon dioxide result from every mole of propane combusted. We had, $(7.51*g)/(44.1*g*mol^-1)$ $=$ $?? "moles of propane"$. So how many moles of...

The equation says that for every $1O_2$, there are $2H_2O$ produced. Therefore, if you have $2.2O_2$, you must have $2.2 * 2H_2O$, or $4.4$ moles of $H_2O$.

$C(s) + O_2(g) rarr CO_2(g)$. The $1:1$ is clear: moles of carbon, and dioxygen gas are equivalent to the moles of carbon dioxide gas generated. We start with $(28.2*g)/(12.01*g*mol)=2.35*mol$...

$"I mol"$ of $"O"_2$ produces $"2 moles"$ of $"H"_2"O"$, so $"2.15 moles"$ of $"O"_2$ produce $2/1×2.15 = "4.30 moles H"_2"O"$

Given: Mass of $"O"_2$; chemical equation (understood) Find: Moles of $"CO"_2$ Strategy: The central part of any problem is to convert moles of something to moles of something else....

$"8.5 mols of gas"$ 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...