How many moles of cesium carbonate will be produced when 5.34 moles of cesium reacts with iron (III) carbonate? (You must write and balance the formula equation!)

2.67 moles of cesium carbonate (Cs2CO3).

Explanation:

We'll begin by writing the equation for the reaction. This is illustrated below:

Cs + Fe2(CO3)3 —> Cs2CO3 + Fe

The above equation can be balance as follow:

There are 2 atoms of Fe on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Fe as shown below:

Cs + Fe2(CO3)3 —> Cs2CO3 + 2Fe

There are 3 atoms of C on the left side and 1 atom on the right side. It can be balance by 3 in front of Cs2CO3 as shown below:

Cs + Fe2(CO3)3 —> 3Cs2CO3 + 2Fe

There are 6 atoms of Cs on the right side and 1 atom on the left. It can be balance by putting 6 in front of Cs as shown below:

6Cs + Fe2(CO3)3 —> 3Cs2CO3 + 2Fe

Thus, the equation is balanced.

From the balanced equation above,

6 moles of Cs reacted with 1 mole of Fe2(CO3)3 to produce 3 moles of Cs2CO3.

Finally, we shall determine the moles of cesium carbonate (Cs2CO3) produced by the reaction of 5.34 moles of cesium (Cs). This can be obtained as follow:

From the balanced equation above,

6 moles of Cs reacted to produce 3 moles of Cs2CO3.

Therefore,

5.34 moles of Cs will react to produce = (5.34 × 3)/6 = 2.67 moles of Cs2CO3.

Thus, 2.67 moles of cesium carbonate (Cs2CO3) were produced from the reaction.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%