Help me in Stoichiometry quiz tomorrow thankz?

Question

help me please to solve this problem..thanks

1.) Tin(IV) chloride is produced in 82.2% yield by the reaction of tin witch chlorine. How much tin is required to produce a kilogram of Tin(IV) chloride.?

Answer 1

nope, the % yield requires you to increase the amount of Sn required to produce 1 Kg of SnCl4, not decrease  the Sn

using molar masses:

? grams Sn = 1000g SnCl4 @  118.7 g/mol Sn / 260.50 g/mol SnCl4 =

455.66  g of Sn is required when there is 100 % yield of SnCL4

@ 82.2 % yield,...  you need more Sn to be added to yield 1 Kg :

455.656 g @ 100% / 82.2%  = 554 grams of Sn

that's your answer: 554 grams of Sn , (aka 0.554 kg)

Answer 2

First, let's write the chemical equation:

Sn + 4Cl -> SnCl4

Now, we can see that one mole of tin will give us one mole of tin (IV) chloride.

First, we need to calculate how many moles of tin chloride we have. The molecular weight of tin (IV) chloride is 260.50 g per mole. One kilogram equals 1,000 grams, so we have:

1,000 / 260.5 = 3.839 moles of Tin Chloride

Now, theoretically, 3.839 moles of tin will give us 3.839 moles of tin chloride. However, in this case (and most real life cases), we only get a percentage of that. Here, only 82.2% of the starting materials are converted to product. Thus:

3.839 / 0.822 = 4.67 moles of Tin

Now, we can convert that to mass by using the molecular weight of tin (118.71 g/mol).

4.67 * 118.71 =  554.38 grams = 0.55 kilograms of Tin

And you're done!