Final answer:
To find the grams of the excess reactant used, first determine the limiting reactant by comparing the amount of product expected for complete reaction of each reactant. Then, calculate the grams of the excess reactant that will be used. The answer is option b: 36.5 grams of the excess reactant will be used.
Step-by-step explanation:
To determine the grams of the excess reactant used, we need to find the limiting reactant first. We can do this by comparing the amount of product expected for complete reaction of each reactant. The balanced equation for the reaction is:
Cr2O3 + 3Si → 2Cr + 3SiO2
From the equation, we can see that 1 mole of Cr2O3 reacts with 3 moles of Si. So, if we start with 145.0 grams of Cr2O3, we can convert it to moles using its molar mass:
(145.0 g Cr2O3) / (151.99 g/mol Cr2O3) = 0.955 mol Cr2O3
Now, we can use the stoichiometry of the balanced equation to find the moles of Si that would react with 0.955 mol Cr2O3:
(0.955 mol Cr2O3) × (3 mol Si / 1 mol Cr2O3) = 2.865 mol Si
Since we have 117.0 grams of Si, we can convert it to moles using its molar mass:
(117.0 g Si) / (28.09 g/mol Si) = 4.163 mol Si
Comparing the moles of Si calculated above with the moles needed for complete reaction, we can see that 2.865 mol Si is the limiting reactant because it is smaller. Now, we can calculate the grams of the excess reactant that will be used:
(4.163 mol Si) - (2.865 mol Si) = 1.298 mol Si
To convert moles of Si to grams of the excess reactant:
(1.298 mol Si) × (28.09 g/mol Si) = 36.47 grams of Si (rounded to the nearest hundredth)
Therefore, the answer is option b: 36.5 grams of the excess reactant will be used.