Page 5 of S 5 10. Complete and balance the following chemical reactions. White f

Question

Page 5 of S 5 10. Complete and balance the following chemical reactions. White formulas first and then balance the equation! Acid-Base: Complete neutralization to form a salt and water. NaOH HSO NH-Cl + Ca(OH)2 ? HNO,+Fe(OH), Redox: Transfer of electrons! CuSO, + Al (Cu metal is one product of this reaction) Al + HCI(hydrogen gas is one product of this reaction) CHo+ O, (combustion is also a redox reaction !) Precipitation: Formation of insoluble salts like AgCl and Pbl, often drives reactions to completion. AgNO, CaCl,> 11. Determine the mass of rust (Fe,O) formed if 2.5 L of oxygen gas (measured at 650 mm Hg and 20 °C) reaction with 8.55 g iron. 4 Fe 30, 2 Fe,O, (HINT: use PV-nRT to determine the number of mols of O, gas and then do the limiting reagent problem.)

Explanation / Answer

10.

1) NaOH(aq) + H2SO4(aq) --------- Na2SO4(aq) + H2O(l)

2 NaOH(aq)   + H2SO4(aq) ----------- Na2SO4(aq) + H2O(l)

2) NH4Cl(aq) + Ca(OH)2 --------------- CaCl2(aq) + NH3(g) + H2O(l)

2 NH4Cl(aq) + Ca(OH)2 (aq) ---------------- CaCl2(aq) + 2 NH3(g) + 2 H2O(l)

3)    HNO3(aq) + Fe(OH)3(aq) -------------- Fe(NO3)3(aq) + H2O(l)

        3 HNO3(aq) + Fe(OH)3(aq) -------------- Fe(NO3)3(aq) +3 H2O(l)

Redox reactions

1)

2 Ca(s) + O2(g) ----------- 2 CaO(s)

2)

3 CuSO4(aq) +2 Al(s) -------3 Cu(s) + Al2(SO4)3(aq)

3)

2Al(s) + 6 HCl(aq) ----------------- 2 AlCl3(aq) + 3 H2(g)

4)

2C4H10(g) +13 O2(g) ------------------ 8CO2(g) + 10 H2O(g)

Precipitation reactions

Pb(NO3)2(aq) + 2 KI(aq) ---------------- PbI2(s) + 2 KNO3(aq)

2 AgNO3(aq) + CaCl2(aq) ----------------- 2 AgCl(s) + Ca(NO3)2(aq)

11)

4 Fe(s) + 3 O2(g) ------------- 2 Fe2O3(s)

mass of Fe = 8.55 gram/mole

molar mass of Fe= 55.8 gram/mole

number of moles of Fe = 8.55/55.8=0.153 moles

V= 2.5L

P=650mm= 650/760 = 0.855 atm

T=20C= 20+273= 293K

R= 0.0821 L-atm/mol-k

PV=nRT

n=PV/RT

n= 0.855x2.5/0.0821x293

n=0.0889 moles

according to equation

3 moles of O2 = 4 moles of Fe

0.0889 moles of O2 =?                   

                                     = 0.0889x4/3=0.118 moles of Fe

number of moles of Fe = 0.118 moles

we need 0.118 moles of Fe. but we have0.153 moles of Fe. so Fe is the excess reagent

Hence O2 is limiting reagent

according to equation

3 moles of O2= 2 moleso f Fe2O3

0.0889 moles of O2 = ?                   
                            = 2x0.0889/3=0.0593moles o f Fe2O3

number of moles of Fe2O3 formed = 0.0593 moles

molar massof Fe2O3 =159.69 gram/mole

massof 0.0593 moles of Fe2O3 = 0.0593 x 159.69=9.469grams

mass of Fe2O3 = 9.47 grams.

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