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Balanced equation
Multiply the given moles of iron by ratio between oxygen and iron from the balanced equation, with moles oxygen in the numerator.
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A student records the mass of a piece of iron. The iron is then left outside until it rusts. What type of reaction occurs, and how does the mass of the object after rusting compare with its original mass?
Iron 26 is not stable in its ground state of $1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^6$ The valance electrons $4s^2 3d^6$ can become somewhat stable by losing...
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Aluminum will react with sulfuric acid in the following reaction: $2Al(s) + 3H_2SO_4(l) -> Al_2(SO_4)_3(aq) + 3H_2(g)$. How many moles of $H_2SO_4$ will react with 18 mol Al? How many moles of each product will be produced?
With 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...
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When iron rusts in air, iron (III) oxide is produced. How many moles of oxygen react with 12.9 mol of iron in the rusting reaction? $4Fe(s) + 3O_2(g) -> 2Fe_2O_3(s)$?
We 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" \...
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How many moles of iron(Ill) sulfide, $Fe_2S_3$, would be produced from the complete reaction of 449g iron(III) bromide in the reaction $2FeBr_3 + 3Na_2S -> Fe_2S_3 + 6NaBr$?
$2FeBr_3 + 2Na_2S rarr Fe_2S_3 + 6NaBr$ ....the which tells you that treatment of approx. $600*g$ $"ferric bromide"$ upon treatment with approx. $160*g$ $"sodium sulfide"$ to give $208*g$ $"ferric sulfide"$,...
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How many moles of $"iron(Ill) sulfide"$, $Fe_2S_3$, would be produced from the complete reaction of $449*g$ $"iron(III) bromide"$ in the reaction $2FeBr_3 + 3Na_2S -> Fe_2S_3 + 6NaBr$?
And thus we simply calculate the moles of ferric bromide. $"Moles of ferric bromide"=(449*g)/(295.96*g*mol^-1)=1.517*mol$. And thus we gets an half equiv of the sulfide salt....which represents a mass of.... $1.517*cancel(mol)xx1/2xx207.90*g*cancel(mol^-1)~=150*g$....
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Iron can be extracted from iron(III) oxide by heating it with coke (carbon). The other product in the extraction reaction is carbon dioxide.
What mass of iron is produced by reacting 550g of iron(III) oxide with 60g of coke?
Balanced Equation $"2Fe"_2"O"_3 + "3C"$$rarr$$"4Fe + 3CO"_2$ This is a limiting reactant question. The maximum amount of $"Fe"$ that can be produced is determined by the limiting reactant. We have...
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If iron (II) oxide gas decomposes into iron liquid and oxygen gas and 6 moles of iron (II) oxide gas are used, how of liters of oxygen gas are produced?
This 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...
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If oxygen is removed from a sample of air as iron rusts, what happens to the partial pressure of oxygen in the air?
There are 2 key things to know 1, Rusting is where oxygen binds to iron and forms iron oxide. So once iron rusts, there is oxygen, just not in air;...
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Iron rusts according to the following reaction: $4Fe (s) + 3O_2 (g) + 4H_2O -> 2Fe_2O_3(H_2O)_2$. If 1.4 liters of oxygen were added to excess water and iron at a pressure of 0.97 atms and a temperature of 55 C, how many grams of iron would rust?
Let us readdress it....iron metal is oxidized to $Fe_2O_3$... $2Fe(s) +3H_2O(l) rarr Fe_2O_3+6H^+ +6e^(-)$ $(i)$ Dioxygen is reduced to water... $1/2O_2 + 2H^+ +2e^(-) rarr H_2O(l)$ $(ii)$ And $1xx(i)+3xx(ii)$ gives.......
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Iron rusts when exposed to oxygen. Iron(II) is produced in the reaction. How many moles of oxygen oxide will react with 2.4 mol of iron in the rusting reaction?
$Fe(s) + 1/2O_2 rarr FeO(s)$ 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...
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