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
For an approximate result, multiply the volume value by 3.785
Answer ≈ 56.7812
Answer:
2Fe + 6HC2H3O2 → 2Fe(C2H3O2)3 + 3H2
Explanation:
There you go
Answer:
In 1 mol of Pb₃(PO₄)₄ occupies 1001.48 grams
Explanation:
This compound is the lead (IV) phosphate.
Grams that occupy 1 mole, means the molar mass of the compound
Pb = 207.2 .3 = 621.6 g/m
P = 30.97 .4 = 123.88 g/m
O = (16 . 4) . 4 = 256 g/m
621.6 g/m + 123.88 g/m + 256 g/m = 1001.48 g/m
Answer:
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Explanation:
