The toxic pigment called whte lead pb3(oh)2(co33)2 has been replaced in white paints by rutile, tio2. how much rutile can be prepared from 379 g of an ore that contains 88.3% ilemnite (fetio3) by mass?
2ftio3 + 4hcl + cl2 ? 2fecl3 + 2tio2 + 2h2o

Answer : The mass of prepared can be 176.48 grams.

Explanation :

First we have to calculate the mass of .

As we are given that the 88.3 % by mass. That means, 88.3 grams of present in 100 grams of solution.

Now we have to calculate the moles of .

Molar mass of = 151.7 g/mole

Now we have to calculate the moles of .

The balanced chemical reaction is,

From the balanced reaction, we conclude that

As, 2 moles of react to give 2 moles of

So, 2.206 moles of react to give 2.206 moles of

Now we have to calculate the mass of .

Molar mass of = 80 g/mole

Therefore, the mass of prepared can be 176.48 grams.

"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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