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
CH4 : H2O
1 : 2
number of moles of H2O = 1.00 x 2
number of moles of H2O = 2.00mol
mass = number of moles x molar mass
mass of H2O = 2.00 x (1 + 1 + 16)
mass of H2O = 36g
Molality
is one way of expressing concentration of a solute in a solution. It is expressed
as the mole of solute per kilogram of the solvent. To calculate for the
molality of the given solution, we need to convert the mass of solute into
moles and divide it to the mass of the solvent.
<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>
<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality
= 0.7543 mol / kg</span>
<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is
0.7453 molal.</span>
Precipitation occurs when the product of the ion concentration exceeds the Ksp.
