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olga55 [171]
3 years ago
5

If 15.00 mL of 0.0100 M Ca(IO3)2 solution are mixed with 0.500 g KI, what is the theoretical yield (in grams) of I2?

Chemistry
1 answer:
n200080 [17]3 years ago
8 0

The theoretical yield of I2 in the reaction would be 0.23 g

<h3>Theoretical yield</h3>

This refers to the stoichiometric yield of a reaction.

From the equation of the reaction:

Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O

The mole ratio of Ca(IO3)2 and I2 is 1: 6

Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100

                                                              = 0.00015 mole

Equivalent mole of I2 = 0.00015 x 6

                                      = 0.009 mole

mass of 0.0009 I2 = 0.0009 x 253.809

                                = 0.23 g

More on stoichiometric calculations can be found here: brainly.com/question/6907332

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4 0
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How can we find the number of moles in equilibrium?
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5 0
3 years ago
A student mixes a solution containing 10.0 g bacl2 (m = 208.2) with a solution containing 10.0 g na2so4 (m = 142.1) and obtains
professor190 [17]
The balanced equation for the above reaction is as follows;
Na₂SO₄ + BaCl₂ --> BaSO₄ + 2NaCl
Na₂SO₄ reacts with BaCl₂ in the molar ratio 1:1
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Number of BaCl₂ moles - 10.0 g / 208.2 g/mol = 0.0480 mol
this means that 0.0480 mol of each reactant is used up, BaCl₂ is the limiting reactant and Na₂SO₄ has been provided in excess. 
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number of BaSO₄ moles formed - 0.0480 mol
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percent yield - 12.0 g/ 11.2 g x 100% = 107% 
this is due to impurities present in the product or product could be wet.
4 0
3 years ago
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