1. For the balanced equation shown below, if 40.2 grams of Fe were reacted with 9.81 grams of H2O, how many grams of H2 would be produced?
3Fe+4H2O=>Fe3O4+4H2 

2. For the balanced equation shown below, if 1.75 grams of P4 were reacted with 1.68 grams of O2, how many grams of P4O10 would be produced?
P4+5O2=>P4O10

P= 31*4= 124g/moles
O=16*2=32g/moles
Moles of P4= 1.75/124= .014 moles
Moles of O2= 1.68/32= .053 moles
Mole Ratio= 1:5
Proportion=
(1/5)*(.014/.053)
                  P4= .053
                  O2= .07= Limiting Reagent
Second Ratio= 5:1= 1:1/5
                P4= 124g/mole
                O10= 160g/mole
                        +________= 284g/mole
Say= .053 moles of O2 -- > .053(1/5)moles of P4O10
                                          .0106 moles of P4O10
  Convert to grams=
                x grams/(284g/mole)= .0106
                x= 3.01 grams

3. For the balanced equation shown below, if 55.4 grams of C2H3F were reacted with 55.6 grams of O2, how many grams of H2O would be produced?
2C2H3F+5O2=>4CO2+2H2O+2HF

C2H3F= 36g/mole
O2= 32g/mole
Moles of C2H3F= 1.54
Moles of O2= 1.74
Mole Ratio: 1:5
Limiting Reagent= O2
Second Ratio= 5:2 => 1:2/5
Molar Mass of H2O= 18g/mole
Final equation=
            1.74(2/5)= .7
           .7(18)=12.6grams

4. For the balanced equation shown below, if 29.3grams of Ba were reacted with 4.98 moles of 02, how many grams of BaO would be produced?
2Ba+O2=> 2BaO

Ba= 137 g/mole
O2= 32 g/mole
Moles of Ba= .214
Moles of O2= .156
Mole Ratio= 2:1
Limiting Reagent= Ba
Second Ratio= 2:2
Molar Mass of BaO= 153g/mole
Final Equation=
            .214(2/2)= .214
            .214(153)= 32.7grams

5. For the balanced equation shown below, if 42.1 grams of ZnS were reacted with 10.9 grams of O2, how many grams of SO2 would be produced?
 2ZnS+3O2=>2ZnO+2SO2

ZnS= 97g/mole
O2= 32g/mole
Moles of ZnS= .43
Moles of O2= .341
Mole Ratio= 2:3
Limiting Reagent= O2
Second Ratio= 3:2=> 1:2/3
Molar Mass of SO2= 64g/mole
Final Equation=
          .341(2/3)= .23
          .23(64)= 14.55

6. For the balanced equation shown below, if 36.8 grams of CBr4 were reacted with 4.56 grams of O2, how many grams of Br2 would be produced?  
CBr4+O2=>CO2+2Br2

CBr4=332 g/mole
O2= 32g/mole
Moles of CBr4= .11
Moles of O2= .143
Limiting Reagent = CBr4
Second Ratio= 1:2
Molar Mass of Br2= 160g/mole
Final Equation=
          .11(2)=.22 
          .22(160)= 35.5grams

7.  For the balanced equation shown below, if 13.2 grams of Fe were reacted with 31.5 grams of Cl2, how many grams of FeCl3 would be produced?
2Fe+3Cl2=>2FeCl3

Fe= 56 g/mole
Cl2= 71 g/mole
Moles of Fe=.24
Moles of Cl2= .443
Mole Ratio= 2:3
Limiting Reagent= Fe
Second Ratio= 2:3=> 1
Molar Mass of FeCl3=162.5 g/mole
Final Equation=
       .24(1)=.24
       .24(162.5)=39grams

8.  For the balanced equation shown below, if 41.4 grams of Al were reacted with 190 grams of Cr2O3, how many grams of Cr would be produced?  
2Al+Cr2O3=>Al2O3+2Cr  

Al = 27g/mole
Cr2O3,= 152g/mole
Moles of Al= 1.53
Moles of Cr2O3= 1.25
Mole Ratio=2:1
Limiting reagent = Al
Second Ratiio= 2:2 =>1
Molar Mass of Cr= 52g/mole
Final Equation=
           1.53(1)= 1.53
          1.53(52)=79.6grams

9. For the balanced equation shown below, if 59.3 grams of CH2S were reacted with 80.6 grams of O2, how many grams of SO3 would be produced? 
CH2S+3O2=>CO2+H2O+SO3

CH2S= 46g/mole
O2=32g/mole
Moles of CH2S= 1.29
Moles of O2= 2.51
Mole Ratio= 1:3
Limting Reagent= O2
Second Ratio= 3:1 =>1:1/3
Molar Mass of SO3= 80g/mole
Final Equation=
           2.51(1/3)=.84
          .84(80)= 67grams

10. For the balanced equation shown below, if 11.0 grams of CH3COF were reacted with 4.97 grams of H2O, how many grams of CH3COOH would be produced? 
CH3COF+H2O=>CH3COOH+HF 

CH3COF= 62g/mole
H2O= 18g/mole
Moles of CH3COF= .177
Moles of H2O= .276
Mole Ratio= 1
Limiting Reagent= CH3COF
Second Ratio= 1
Molar Mass of CH3COOH= 69g/mole
Final Equation=
           .177(1)=.177
           .177(69)=12.2grams

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