First, we must know what ferrous sulfate is. Ferrous refers to \(\displaystyle Fe^{2+}\) , and sulfate has the formula \(\displaystyle SO_4^{-2}\). When we combine the two together we get \(\displaystyle FeSO_4\).

Calcium is a divatent cation and iodide is a monovalent anion, so their salt is \(\displaystyle CaI_2\). The ion exchange reaction is then:

\(\displaystyle FeSO_4 + CaI_2 \rightarrow FeI_2 +CaSO_4\)

The net ionic equation is derived by removing all spectator ions from the total ionic equation (in which all ions are listed). To put it another way, the net ionic equation involves only the ions that participate in a reaction which, in this case, is the precipitation of barium sulfate.

Begin by writing all aqueous compounds in their dissociated (ionic) forms.

\(\displaystyle Ba(NO_{3})_{2} (aq) + K_{2}SO_{4} (aq) \rightarrow 2KNO_{3} (aq) + BaSO_{4} (s)\)

\(\displaystyle Ba^{2+}+2NO_3^-+2K^++SO_4^{2-}\rightarrow2K^++2NO_3^-+BaSO_4(s)\)

Cancel out any ions that appear in equal quantities on both sides of the equation. In this case, we can cancel the nitrate and potassium ions.

\(\displaystyle Ba^{2+}+SO_4^{2-}\rightarrow BaSO_4(s)\)

This is our net ionic equation.

Combustion is the chemical reaction of a hydrocarbon with molecular oxygen, and it always produces carbon dioxide and water. Knowing the reactants and products, the unbalanced equation must be: 

\(\displaystyle C_4H_{10} + O_2 \rightarrow CO_2 + H_2O\)

We start by balancing the hydrogens. Since there are 10 on the left and only 2 on the right, we put a coefficient of 5 on water.

\(\displaystyle C_4H_{10} + O_2 \rightarrow CO_2 + 5H_2O\)

Similarly, we balance carbons by putting a 4 on the carbon dioxide.

\(\displaystyle C_4H_{10} + O_2 \rightarrow 4CO_2 + 5H_2O\)

To find the number of oxygens on the right, we multiply the 4 coefficient by the 2 subscript on O (which gets us 8 oxygens) and then add the 5 oxygens from the 5 water molecules to get a total of 13. The needed coefficient for \(\displaystyle O_2\) on the left would then have to be 13/2.

\(\displaystyle C_4H_{10} + \frac{13}{2}O_2 \rightarrow 4CO_2 + 5H_2O\)

Because fractional coefficients are not allowed, we mutiply every coefficient by 2 to find our final reaction:

\(\displaystyle 2C_4H_{10} + 2*\frac{13}{2}O_2 \rightarrow 2*4CO_2 + 2*5H_2O\)

\(\displaystyle 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O\)

When we check the activity series, it is fairly easy to see that aluminum metal is more reactive than zinc metal. So, in this case, the two metals undergo a redox reaction, where the aqueous \(\displaystyle Zn^{2+}\) is reduced to solid \(\displaystyle Zn\), and the solid \(\displaystyle Al\) is oxidized to aqueous \(\displaystyle Al^{3+}\). These charges are the common oxidation states for zinc and aluminum and should be memorized.

Because \(\displaystyle Al^{3+}\) is the new species, it bonds with 3 \(\displaystyle Cl^-\) ions. The unbalanced equation is:

\(\displaystyle Al(s) + ZnCl_2(aq) \rightarrow AlCl_3 (aq) + Zn (s)\)

We note that there are 2 chlorine atoms on the left and 3 chlorine atoms on the right. To balance, we use a 3 coefficient on the left and a 2 coefficient on the right. This gives a total of 6 chlorine atoms on eahc side.

\(\displaystyle Al(s) + {\color{Red} 3}ZnCl_{\color{Blue} 2}(aq) \rightarrow{\color{Blue} 2}AlCl_{\color{Red} 3} (aq) + Zn (s)\)

However, now we have also increased the amounts of zinc and aluminum. We copy the necessary coefficients to balance those—2 for aluminum on the left, 3 for zinc on the right—and we are done:

\(\displaystyle 2Al(s) + 3ZnCl_2(aq) \rightarrow 2AlCl_3 (aq) + 3Zn (s)\)

In the unbalanced equation from the question, the reactant side contains two hydrogen atoms and one oxygen atom. On the product side of the equation, there are four hydrogens and two oxygen atoms. So for every oxygen atom there are two hydrogen atoms giving a ratio of \(\displaystyle 1:2\) (oxygen : hydrogen). Putting a coefficient of two in front of the water molecule on the reactant side of the equation balances the equation. Therefore, there are four hydrogen and two oxygen atoms on both sides of the balanced chemical equation.

We must first calculate the molecular weight of \(\displaystyle Na_{2}SO_{4}\) to use as a conversion factor:

\(\displaystyle MW_{Na_{2}SO_{4}}=(\frac{23g}{mole}\times 2)+(\frac{32g}{mole}\times 1)+(\frac{16g}{mole})=142\frac{g}{mole}\)\(\displaystyle 15.5g \times \frac{mole}{142g}=0.1092\ moles\ Na_{2}SO_{4}\)

For every mole of \(\displaystyle Na_{2}SO_{4}\) reacted, one mole of \(\displaystyle BaSO_{4}\) is produced.

Therefore, 0.1092 moles of \(\displaystyle BaSO_{4}\) is produced.

We must first calculate the molecular weight of \(\displaystyle Na_{2}SO_{4}\) to use as a conversion factor:

\(\displaystyle MW_{Na_{2}SO_{4}}=(\frac{23g}{mole}\times 2)+(\frac{32g}{mole}\times 1)+(\frac{16g}{mole})=142\frac{g}{mole}\)\(\displaystyle 15.5g \times \frac{mole}{142g}=0.1092\ moles\ Na_{2}SO_{4}\)

For every 1 mole of \(\displaystyle Na_{2}SO_{4}\) reacted, 2 moles of \(\displaystyle NaCl\) is produced.

Therefore,the number of moles of \(\displaystyle NaCl\) produced will be double the number of moles of \(\displaystyle Na_{2}SO_{4}\) used up in the reaction:

\(\displaystyle Moles\ of\ NaCl\ produced = 0.1092\ moles\ Na_{2}SO_{4}\times2=0.2183\ moles\)

The number of atoms on the reactant side of the chemical equation are:

\(\displaystyle Total\ O=6,\ Total\ P=1,\ Total\ H=5,\ Total\ Ca=1\)

The number of atoms on the product side of the chemical equation are:

\(\displaystyle Total\ O=5,\ Total\ P=1,\ Total\ H=3,\ Total\ Ca=1\)

In a balanced chemical equation, the total number of atoms of each element must be equal on the reactant and product side of the chemical equation. There is one more oxygen and two more hydrogen atoms on the reactant side of the chemical equation. Giving the water molecule on the product side of the chemical equation would balance the chemical equation so that both sides contain six oxygens and five hydrogens:\(\displaystyle H_3PO_4+Ca(OH)_2\rightarrow 2H_2O+CaHPO_4\)

The process of balancing a chemical equation is called "balancing by inspection" which is done by trial and error. The easiest way to approach the problem given is by balancing the species containing an element with the most complicated formula occurring in one reactant and one product. In this case, we can either begin by balancing the carbon or hydrogen in the propane molecule. Let's start with the carbon by adding a coefficient of 3 to the carbon dioxide molecule on the product side of the equation which gives:

\(\displaystyle C_{3}H_{8}\ +\ O_{2}\rightarrow\ 3CO_{2}\ +\ H_{2}O\)

Let's now balance the hydrogens by adding a coefficient of 4 to the product side of the equation:

\(\displaystyle C_{3}H_{8}\ +\ O_{2}\rightarrow\ 3CO_{2}\ +\ 4H_{2}O\)

Lastly, we can balance the number of oxygens in the reaction:

\(\displaystyle C_{3}H_{8}\ +\ 5O_{2}\rightarrow\ 3CO_{2}\ +\ 4H_{2}O\) 

Based on the molecular equation, for every 1 mole of \(\displaystyle NH_{3}\) reacted, 1 mole of \(\displaystyle NH_{4}Cl\) is formed as a product. Using dimensional analysis, we can use this relationship to determine the moles of \(\displaystyle NH_{3}\) reacted:

\(\displaystyle 0.150\ moles\ of\ NH_{3}\times\frac{1\ mole\ NH_{4}Cl}{1\ mole\ NH_{3}}=0.150\ moles\ of\ NH_{4}Cl\)

We must calculate the molecular weight of \(\displaystyle NH_{4}Cl\):

\(\displaystyle MW_{NH_{4}Cl}=(14\frac{g}{mole}\times 1)+(1\frac{g}{mole}\times 4)+(35\frac{g}{mole}\times1)=53\frac{g}{mole}\)

To calculate the number of grams of \(\displaystyle NH_{4}Cl\), we need to convert the number of moles of \(\displaystyle NH_{4}Cl\) to grams using the molecular weight as a conversion factor.

\(\displaystyle 0.150\ moles\ of\ NH_{4}Cl\ \times 53\frac{g}{mole}=7.95\ g\)