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2 April, 20:51

What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of cr2o3 with 8.00 g of aluminum according to the chemical equation below? 2al + cr2o3 - -> al2o3 + 2cr?

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  1. 2 April, 21:07
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    The theoretical yield of chromium will be 15.05 g

    Solution:

    The balance chemical equation

    2Al + Cr2o3 - -> Al2o3 + 2Cr

    From given mass, we can calculate moles

    mass of Al = 8 g mass of Cr2O3 = 40g

    Molar mass of Al = 26.98g/mol Molar mass of Cr2O3 = 151.9g/mol

    Mole of Al = 8/26.98 Mole of Cr2O3 = 40/151.9

    Mole of Al = 0.29 mole Mole of Cr2O3 = 0.26 mol

    Here, we have to find out the limiting reactant (The reactant which will give the least amount of product) so from balance chemical equations

    No. of mole of Al = No. of moles of Cr

    2 = 2

    0.29 = 2/2 * 0.29 = 0.29 mole

    No. of mole of Cr2O3 = No. of moles of Cr

    1 = 2

    0.26 = 2/1 * 0.26 = 0.52 mole

    So we can say, Al is a limiting reactant so

    No. of mole of Al = No. of moles of Cr

    2 = 2

    0.29 = 2/2 * 0.29 = 0.29 mole

    Moles of Cr = 0.29 mol

    Molar mass of Cr = 51.9 g/mol

    mass of Cr = 0.29 * 51.9 = 15.05 g

    Conclusion:

    The calculated mass from balance chemical equation of Cr will be 15.05 g.
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