60g of solid Zn are added to 50mL of a 0.15M solution of HCl and undergo a chemi

Question

60g of solid Zn are added to 50mL of a 0.15M solution of HCl and undergo a chemical reaction according to the following equation: Zn(s) + 2 HCl(aq) ? ZnCl2(aq) + H2(g) a) How many grams of ZnCl2 can be produced from this reaction? b) How would you classify this reaction? As acid/base, oxidation/reduction or precipitation? c) Assuming that all the H2 that is produced in the reaction is taken to a different container, how much volume would it occupy under STP conditions? d) If the produced H2 is later mixed with 50g of F2 in balloon. What would be the partial pressure of F2 is the balloon is under 3atm of pressure? e) Pretend that the HCl of this reaction is used in the neutralization of 45mL of a solution of Ga(OH)3. What was the molarity of the Ga(OH)3 solution?

Explanation / Answer

50mL/ 1000= 0.05L to find the moles of HCl i multiply mol=M x V 0.15 x 0.05= 0.0075 mol HCl then to find the mole of ZnCl 0.0075 mol HCl ( 1 mole ZnCl / 2 mole HCl) = 0.0375 mol of ZnCl then from that i converted the moles of ZnCl Zn= 65 Cl= 2 x 35 65+70=135g/mol 0.00375 mol ZnCl (135g / 1mol) = 0.50625g ZnCl

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