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
- Elliot YangLv 41 decade agoFavorite 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
- imapottermanLv 41 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.
- JoanneLv 44 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.