The answer relates to the stoichiometric equation, which tells us that $159.7*g$$Fe_2O_3$ reacts with $84.0*g$$CO$ to give $112*g$$Fe$.
You have $150*"lbs"xx454*g*"lb"^-1$$=$$68,100*g$$Fe_2O_3$. Given the equations you should be able to tells us the quantities of steel produced, and the quantity of gas required. If you have difficulties, state them here, and someone will help you.
And this is the effectual chemical reaction of our civilization. If you ever have a chance to tour a blast furnace (and most are in China and India now) do...
A substance's essentially tells you how many grams of each constituent element you get per $"100 g"$ of said substance. In this case, a percent composition of $69.94%$...
$Fe_2O_3$ has a molar mass of $159.69*g*mol^-1$.. We work from a molar quantity with respect to iron of $(58.7*g)/(55.8*g*mol^-1)=1.05*mol$, and thus with respect to oxygen, there are is $1.05*molxx3/2xx16*g*mol^-1=25.25*g$ mass....
As can be seen from the chemical equation, each mole of $Fe_2CO_3$ needs 3 moles of $CO$. If 30 moles of $CO$ are used, only 10 moles of $Fe_2CO_3$ can...
The equation for the formation of coal gas is $"2C(s)" + "2H"_2"O(g)" → "CH"_4("g") + "CO"_2("g")$ This is our target equation. We must create it from equations (1) - (3)....
$"Moles of ferric oxide "=(0.18*g)/(159.69*g*mol^-1)$ $=$ $1.13xx10^-3*mol" metal oxide"$. $"Moles of carbon monoxide "=(0.11*g)/(28.0*g*mol^-1)$ $=$ $3.93xx10^-3*mol" CO"$. $Fe_2O_3$ is the reagent in deficiency (why?), and thus $2xx1.13xx10^-3*molxx55.85*g*mol^-1~=0.150*g$ iron metal are...
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...
$"Moles of ferric oxide"$ $=$ $(100*cancel(kg)xx10^3*cancelg*cancel(kg^-1))/(159.69*cancelg*mol^-1)$ $=$ $626.2*mol$. And clearly, from the of the reaction, $2xx626.2*cancel(mol)xx55.85*cancelg*cancel(mol^-1)xx10^-3kg*cancel(g^-1)$ $=$ $??$
Since $950$ $"g"$ is the desired mass of $"CS"_2$, and the is $86%$, the theoretical yield of $"CS"_2$ is $950$ $"g CS"_2"$ $xx 100/86 = ul(1107color(white)(l)"g CS"_2$ Now, let's...
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...
