CHEMISTRY PROBLEM CHEMISTRY PROBLEM?

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

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  • 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

    You've had a couple of good answers already but have you considered that the total depth of your tank is probably much less that the upper levels of Arowanas native waters. It is perfectly normal for them to use the upper three feet - all of a normal arowana tank. I weaned my Asian off live food onto pellets by throwing a few crickets in at the same time. BTW - it should be paternal care, fraternal involves brothers not fathers.

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