I need help completing and balancing equations Mg+O2 --> KOH + NHO3 --> Febr3 + KOH--> CuSO4 + Na--> C6H14+O2 -->   smallest set of whole number coefficients and...

I need help completing and balancing equations

Mg+O2 -->

KOH + NHO3 -->

Febr3 + KOH-->

CuSO4 + Na-->

C6H14+O2 -->

 



smallest set of whole number coefficients and classify as a single replacement, etc

Expert Answers
ndnordic eNotes educator| Certified Educator

First reaction is synthesis:

2Mg + O2 --> 2MgO

Second reaction is double replacement or neutralization

KOH + HNO3 -->  KNO3  + HOH

Third reaction would also be double replacement

FeBr3 + 3 KOH --> Fe(OH)3  + 3 KBr

Fourth reaction is single replacement:

CuSO4 + 2 Na  -->  Na2SO4  + Cu

Last reaction is combustion:

2 C6H14  +   19 O2  -->  12 CO2  + 14 HOH

givingiswinning | Student

2Mg + O2 => 2MgO

KOH + HNO3 =>  KNO3  + HOH

FeBr3 + 3KOH => Fe(OH)3  + 3KBr

CuSO4 + 2 Na  =>  Na2SO4  + Cu

2C6H14  +   19O2  =>  12CO2  + 14HOH

 
 
atyourservice | Student

2Mg + O2 -> 2MgO

KOH + HNO3 ->  KNO3  + HOH

FeBr3 + 3KOH -> Fe(OH)3  + 3KBr

CuSO4 + 2 Na  ->  Na2SO4  + Cu

2C6H14  +   19O2  ->  12CO2  + 14HOH

Jyotsana | Student

Synthesis

2Mg+O2=>2MgO

Double Replacement

KOH+HNO3=>KNO3+HOH

Double Replacement

FeBr3+3KOH=>Fe(OH)3+3KBr

Single Replacement

CuSO4+2Na=>Na2S04+Cu

Combustion

2C6H14+19O2=>12CO2+14HOH

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