Finish balancing the following equation:
_al(s) + 3zn(no3)2(aq) → __ai(no3)3(aq) +_zn(s)
what coefficient does zn(s) have in the balanced equation?

3 is coefficient of Zn in balanced equation

Explanation:

Balanced Equation:

2Al(s) + 3Zn(NO3)2(aq) → 2AI(NO3)3(aq) + 3Zn(s)

The balance equation is:

2 Al(s) + 3 Zn(NO₃)₂(aq) → 2 AI(NO₃)₃(aq) + 3 Zn(s)

Zn (s) has the coefficient 3 in the balanced equation.

Explanation:

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

For balance the chemical equation, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts.

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.  

So, in this case:

__Al(s) + 3 Zn(NO₃)₂(aq) → __AI(NO₃)₃(aq) + __Zn(s)

Left side: 1 Al, 3 Zn, 6 (3*2) N and 18 (3*3*2) O

Right side: 1 Al, 1 Zn, 3 N and 9 (3*3) O

__Al(s) + 3 Zn(NO₃)₂(aq) → 2 AI(NO₃)₃(aq) + __Zn(s)

Left side: 1 Al, 3 Zn, 6 (3*2) N and 18 (3*3*2) O

Right side: 2 Al, 1 Zn, 6 (2*3) N and 18 (3*3*2) O

2 Al(s) + 3 Zn(NO₃)₂(aq) → 2 AI(NO₃)₃(aq) + __Zn(s)

Left side: 2 Al, 3 Zn, 6  N and 18 O

Right side: 2 Al, 1 Zn, 6 N and 18  O

2 Al(s) + 3 Zn(NO₃)₂(aq) → 2 AI(NO₃)₃(aq) + 3 Zn(s)

Left side: 2 Al, 3 Zn, 6  N and 18  O

Right side: 2 Al, 3 Zn, 6 N and 18  O

You can see that you have the same amount of elements on each side of the reaction. Then the equation is balanced.

Zn (s) has the coefficient 3 in the balanced equation.

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.