The reaction is:
So, ratio between sodium and sodium oxide is
Therefore, if ten moles of sodium are used, then we produce:
The overall reactions look like this $2"Al"_2"O"_text(3(l]) -> 4"Al"_text((l]) + 3"O"_text(2(g]) uarr$ and $2"Na"^(+)"Cl"^(-) -> 2"Na"_text((s]) + "Cl"_text(2(g]) uarr$ Now, the half-reactions that are of...
1 Answers 1 viewsWith reference to the given balanced equation for the , the needed number of moles for each remaining compound involved in the reaction using the mass of $Al$ as basis...
1 Answers 1 viewsBalanced equation $"4Fe(s) + 3O"_2("g")"$$rarr$$"2Fe"_2"O"_3("s")"$ Multiply the given moles of iron by ratio between oxygen and iron from the balanced equation, with moles oxygen in the numerator. $6.4color(red)cancel(color(black)("mol Fe"))xx(3"mol O"_2)/(4color(red)cancel(color(black)("mol...
1 Answers 1 viewsWe have the balanced equation: $4Fe(s)+3O_2(g)->2Fe_2O_3(s)$ And so, $4$ moles of iron react with $3$ moles of oxygen, and therefore, $12.9 \ "mol"$ of iron would react with: $12.9color(red)cancelcolor(black)("mol" \...
1 Answers 1 viewsThe molecular mass of natrium oxide is $61.98$ $g*mol^(-1)$. If $5$ $mol$ natrium react, then $5/2$ $molxx61.98$ $g*mol^(-1)$ $=$ $154.95$ $g$ natrium oxide should result. So what have I...
1 Answers 1 viewsYou know that $2"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))$ and that the reaction produced $"234 g"$ of sodium chloride. Convert this to moles by using the...
1 Answers 1 viewsLet us first look at the equation. As per the equation , 2 moles of Magnesium produces two moles of MgO . If we double the amount of Magnesium ,...
1 Answers 1 viewsYou need to start by writing the balanced chemical equation that describes the of sodium chloride $color(red)(2)"Na"_ ((s)) + "Cl"_ (2(g)) -> 2"NaCl"_ ((s))$ The balanced chemical...
1 Answers 1 viewsThis means that for $6mol$ of $FeO$, only $3mol O_2$ is produced (half the moles). The volume of the oxygen is dependant on temperature and pressure. E.g. : If...
1 Answers 1 views$Fe(s) + 1/2O_2 rarr FeO(s)$ This is a very simplified representation of the oxidation of . But if there are $2.4*mol$ of iron metal, it should react with $1.2*mol$ of...
1 Answers 1 views