How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following chemical equation?

MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O

a. 0.82 g

b. 5.8 g

c. 13.0 g

d. 14.6 g

e. 58.4 g

3 Answers

  • 1 decade ago
    Favorite Answer

    number of moles of MnO2 = 16.0/(32+55) = 0.184 moles

    number of moles of HCl = 30.0/(1+35.5) = 0.822 moles

    MnO2 is limiting reagent

    so grams of Cl2 can prepared = 0.184 * 71 = 13.0g

    so answer is c

  • 1 decade ago

    The GMW of hydogen is 1 so that means that there is 29 g of Cl in the given amount of HCl. You notice that half of the Cl is used in MnCl2 so, therefore, half of the Cl is in the gas Cl2. Your answer is d) 14.6g.

  • Joanne
    Lv 4
    4 years ago

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