Question

Determine the limiting reactant when a mixture of `"5000 g"` of `"SiO"_2` and `"5000 g"` of carbon react and how many grams of excess reactant remain after the reaction is complete? Equation of reaction: `"SiO"_2 + 3C"``rarr``"SiC + 2CO"`

Answer 1

Well, you already written the stoichiometric equation....

Explanation:

`underbrace(SiO_2(s) +3C(s))_"96.1 g" rarr`

`underbrace(SiC(s) + 2CO(g)uarr)_"96.1 g"`

And so we address the stoichiometric equivalence....

`"Moles of silicon dioxide"=(5000*g)/(60.08*g*mol^-1)=83.2*mol`

`"Moles of carbon"=(5000*g)/(12.011*g*mol^-1)=416.3*mol`

Now, clearly, there is a vast molar excess of carbon ... as we would expect ... carbon is cheap and readily available...

And at most, we could access `83.2*mol` of silicon carbide. Agreed?

And thus `416.3*mol-3xx83.2*mol=166.7*mol` of carbon would remain after complete reaction. And I will let you calculate the mass of the carbon reductant...