HW 7 me Zn 65.4, CI 35.5, O 16.0, c = Answer the questions in the spaces below a
ID: 548598 • Letter: H
Question
HW 7 me Zn 65.4, CI 35.5, O 16.0, c = Answer the questions in the spaces below and show all working. 12.0 amu 1. Balance the following equations: N2(g) + H2(g) NH3(g) b) NaOH(aq)+HCl(aq) NaCl(aq) + H20(1) c) Pb(NO3)2(aq) + HCl(aq) PbC12(s) + HNO3(aq) 2. Balance the equation below and use it to calculate the moles of Fe formed from 0.50 moles of Fe2O3: FeO(s) + 3CO(g) 2 Fe(s) + 3CO2(g) 3. Balance the equation below then calculate the mass of ZnCl2 by obtained using a 10.0 g of zinc metal: Zn(s) + HCl(aq) ZnCl2(aq) + H2(g)Explanation / Answer
a. N2(g) + 3H2(g) --------> 2NH3(g)
b. NaOH (aq) + HCl(aq) -----------> NaCl(aq) + H2O(l)
c. Pb(NO3)2(aq) + 2HCl(aq) -----------> PbCl2(s) + 2HNO3(aq)
2.
Fe2O3(s) + 3CO(g) ----------> 2Fe(s) +3CO2(g)
1 mole of Fe2O3 react with Co to gives 2 moles of Fe
0.5 moles of Fe2O3 react with CO to gives = 2*0.5/1 = 1 moles of Fe
3. Zn(s) + 2HCl (aq) ---------> ZnCl2(aq) + H2(g)
1 mole of Zn react with HCl to gives 1 moles of ZnCl2
65.4g of Zn react with HCl to gives 136.4g of ZnCl2
10g of Zn react with HCl to gives = 136.4*10/65.4 = 20.86g of ZnCl2
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