How many grams of Cl2 can be prepared from the reaction of 16.0g of MnO_(2) and 30.0g HCl according to the following chemical equation? MnO_(2) + 4 HCI -> MnCI_(2 )+CI_(2) +2H_(2)O
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Expert's answer
2020-10-19T14:06:19-0400
"\\nu(HCl)=30g\/(35.5+1)g\/mol=0.822mol"
"\\nu(MnO_2)=16g\/(55+2*16)g\/mol=0.184mol"
That is, acid in excess. Then, following the equation "\\nu(MnO_2)=\\nu(Cl_2)"
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