Chemistry, 22.03.2020 05:00 Adolfosbaby

Calculate the maximum numbers of moles and grams of iodic acid (HIO3) that can form when 341 g of iodine trichloride reacts with 151.7 g of water:

ICl3 + H2O → ICl + HIO3 + HCl [unbalanced]

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Answer from: sony72

0.73moles

128.48‬g

Explanation:

Given parameters:

Mass of ICl₃ = 341g

Mass of H₂O = 151.7g

Unknown:

Maximum numbers of moles and grams of iodic acid = ?

Solution:

To solve this problem, one needs to obtain a balanced chemical equation for a start;

2ICl₃ + 3H₂O → ICI + HIO₃ + 5HCl

Now, we need to find the limiting reagent because it determines the extent of the reaction;

Number of moles = $\frac{mass}{molar mass} \\$

molar mass of ICl₃ = 127 + 3(35.5) = 233.5g/mol

molar mass of H₂O = 2(1) + 16 = 18g/mol

Number of moles of ICl₃ = $\frac{341}{233.5}$ = 1.46moles

Number of moles of H₂O = $\frac{151.7}{18}$ = 8.43moles

From the balanced equation;

2 moles of ICl₃ combined with 3 moles of water

1.46 moles of ICl₃ will need $\frac{1.46 x 3}{2}$ = 2.19 moles

But we were given 8.43moles of water;

This suggests that water is excess

Now we can relate the known to the unknown:

2 moles of ICl₃ produced 1 mole of HIO₃

then 1.46 mole of ICl₃ will produce $\frac{1.46}{2}$ = 0.73moles

Mass of HIO₃ = number of moles x molar mass

= 0.73moles x (1 + 127 + 3(16))

= 128.48‬g

Answer from: Quest

the awnser is group 12

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³