The reaction of Cr2O3 with silicon metal at high temperatures will make chromium metal.

2Cr2O3(s)+3Si(s) > 4Cr(l)+3SiO2(s)

The reaction is begun with 161.00 g of Si and 139.00 g of Cr2O3.

How many grams of the excess reactant is left after the reaction is complete?

122.9 g of will be left from the given masses of both reactants.

Explanation:

To calculate the moles :

According to stoichiometry :

2 moles of require 3 moles of

Thus 0.91 moles of will require= of

Thus is the limiting reagent as it limits the formation of product and is the excess reagent.

Thus (5.75-1.36) = 4.39 moles  of excess reagent.

Mass of

Thus 122.9 g of will be left from the given masses of both reactants.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

sup bro, im dumb, but this is what i found xd

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