Answer to Question #99123 in General Chemistry for m

Question #99123
Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid,HCl (aq) , as described by the chemical equation

MnO2(s) + 4 HCl(aq)= MnCl2(aq) + 2 H2O(l)+Cl2(g)

How much MnO2 should be added to excess HCl(aq) to obtain 245mL Cl2(g) at 25 °C and 705 Torr ?
1
Expert's answer
2019-11-21T06:26:36-0500

From the combined gas law:


V2(Cl2)=T2p1V1(Cl2)T1p2V_2(Cl_2) = {T_2p_1V_1(Cl_2) \over T_1p_2}

V2(Cl2)=273.15705245298.15760=208mLV_2(Cl_2) = {273.15*705*245 \over 298.15*760} = 208 mL

n(Cl2)=0.20822.4=0.009moln(Cl_2) = {0.208 \over 22.4} = 0.009 mol

From the stoichiometric reaction relationships:


n(MnO2)=n(Cl2)=0.009moln(MnO_2) = n(Cl_2) = 0.009mol

m(MnO2)=nM=0.00986.937=0.782gm(MnO_2) = n*M = 0.009*86.937 = 0.782g

Answer: 0.782 g

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