If 5.00g of iron metal is reacted with 0.950g of Cl2 gas, how many grams of ferric chloride (FeCl3) will form?

1.45g of FeCl3

Explanation:

The equation of the reaction is given as;

2Fe + 3Cl2 --> 2FeCl3

2 mol of Fe reracts with 3 mol of Cl2 to form 2 mol of FeCl3

Upon converting to mass using;

Mass = Number of moles * Molar mass

( 2 * 55.85 = 111.7g ) of Fe reacts with ( 3 * 71 = 213g ) of Cl2 to form ( 2 * 162.2 = 324.4g) of FeCl3

Cl2 is the limiting reactant as it determines how much of FeCl3 is formed

213g of Cl2 = 324.4g of FeCl3

0.950g of Cl2 = x

x = (0.950 * 324.4 ) / 213

x = 1.45g of FeCl3

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"