Lead (II) nitrate reacts with iron (III) chloride to produce lead (II) chloride and iron (III) nitrate. What mass of iron (III) chloride would be needed to make 225 mL of 5.25 M iron (III) chloride solution?

8 carbon atoms forming single bonds with 18 hydrogen atoms; a triple bond exists between the third and fourth carbon atom from the right

explanation:

"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"

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