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The reaction of Cr2O3 with silicon metal at high temperatures will make chromium metal.
2CrO3(s) + 3Si(s)----> 4Cr(l) + 3SiO2 (s)
The reaction is begun with 167.00 g of Si and 146.00 g of Cr2O3. How many grams of the excess reactant are left after the reaction is complete?

I found which one was the L.R but I can't figure out how to find the excess amount of the E.R. Help will be greatly appreciated!!

Respuesta :

Answer:- 126.53 g of Si are left.

Solution:- Let's calculate the moles of any of the product from the given grams of the reactants and see which one gives us less moles, that would be the limiting reactant and the other would be the excess reactant.  After this, we could calculate the used amount of the excess reactant and on subtracting this used amount from total given amount, the excess amount is calculated. The calculations are shown below:

[tex]167.00gSi(\frac{1molSi}{28.085gSi})(\frac{4molCr}{3molSi})[/tex]

= 7.928 mol Cr

Similarly the calculations of moles of Cr from other reactant:

[tex]146.00gCr_2O_3(\frac{1molCr_2O_3}{151.99gCr_2O_3})(\frac{4molCr}{2molCr_2O_3})[/tex]

= 1.921 mol Cr

[tex]Cr_2O_3[/tex] gives the less moles of the product and so it is limiting reactant and hence the excess reactant is Si.

Calculations for the amount of Si used to reactant with limiting reactant:

[tex]146.00gCr_2O_3(\frac{1molCr_2O_3}{151.99gCr_2O_3})(\frac{3molSi}{2molCr_2O_3})(\frac{28.085gSi}{1molSi})[/tex]

= 40.47 g Si

Amount of excess reactant(Si) left = 167.00 g - 40.47 g = 126.53 g

So, after the reaction is completed, 126.53 g of excess reactant(Si) are left.

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