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What is the formal charge on the sulfur atom in sulfur trioxide $SO_3$?
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...
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7.0 moles of sulfur atoms and 10 moles of oxygen molecules are combined to form the maximum amount of sulfur trioxide, how many moles of which reactant remain unused at the end of the reaction?
.........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...
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Solid arsenic reacts with Oxygen gas to synthesize solid arsenic trioxide. How many moles of arsenic trioxide will be produced from 3.4 moles of oxygen gas?
$4As$ + $ 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...
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A mole of sulfur trioxide molecules contains how many oxygen atoms?
Where $N_A-="Avogadro's number"-=6.022xx10^23*mol^-1$. And so if we gots a mole of sulfur trioxide molecules we gots $3*molxx6.022xx10^23*mol^-1$ $"oxygen atoms"$... i.e. $18.066xx10^23*"oxygen atoms.."$
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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
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...
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In a combustion reaction, initially there were $77.8*g$ of oxygen gas, and a $22.8*g$ mass of carbon; $18.0*g$ of oxygen remained after the reaction. If the carbon were completely oxidized, what mass of carbon dioxide resulted?
$77.8*g-18.0*g$ oxygen gas reacted, and $22.8*g$ carbon was consumed. Given complete combustion, CLEARLY, $(59.8+22.8)*g=82.6*g$ $CO_2(g)$ were produced. Why $"CLEARLY?"$. Because mass is conserved in all chemical reactions. $"Garbage in must...
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What volume of $O_2$ (g) at $350^oC$ and a pressure of 5.25 atm is needed to completely convert 5.00 g of sulfur to sulfur trioxide?
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........
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What volume of $O_2(g)$ at 350. $"^o$C and a pressure of 5.25 atm is needed to completely convert 10.0 g of sulfur to sulfur trioxide?
$"Moles of sulfur"$ $=$ $(10.0*g)/(32.06*g*mol^-1)=0.312*mol$. 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: $V=(nRT)/P$ $=$ $(0.468*cancel(mol)xx0.0821*L*cancel(atm)*cancel(K^-1)*cancel(mol^-1)xx623*cancel(K))/(5.25*cancel(atm))$...
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What volume of oxygen is needed to react with solid sulfur to form 6.20 L of sulfur dioxide?
Given that the reaction proceeds quantitatively, whatever the conditions of temperature and pressure, if there were $6.20*L$ of $SO_2(g)$, there were $6.20*L$ $"dioxygen"$ required to make it. demands this....
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If a reaction of sulfur dioxide with dioxygen gas to form sulfur trioxide starts with $"5 mols SO"_2$ and $"5 mols O"_2$, and $60%$ of the $"SO"_2$ was consumed in the reaction, what is the total mols of gas at equilibrium?
$"8.5 mols of gas"$ 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...
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