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IrinaVladis [17]
3 years ago
10

Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO3 and HI are mixed. Give th

e complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO3 and HI are mixed. 2 H (aq) SO32-(aq) → H2O(l) SO2(g) 2 H (aq) SO32-(aq) Mg2 (aq) 2 I-(aq)→ H2SO3(s) MgI2(aq) 2 H (aq) SO32-(aq) → H2SO3(s) Mg2 (aq) 2 I-(aq) → MgI2(s) No reaction occurs.
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0

Answer:

Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)

Explanation:

<em>Give the complete ionic equation for the reaction (if any) that occurs when aqueous solutions of MgSO₃ and HI are mixed.</em>

When MgSO₃ reacts with HI they experience a double displacement reaction, in which the cations and anions of each compound are exchanged, forming H₂SO₃ and MgI₂. At the same time, H₂SO₃ tends to decompose to H₂O and SO₂. The complete molecular equation is:

MgSO₃(aq) + 2 HI(aq) ⇄ MgI₂(aq) + H₂O(l) + SO₂(g)

In the complete ionic equation, species with ionic bonds dissociate into ions.

Mg²⁺(aq) + SO₃²⁻(aq) + 2 H⁺(aq) + 2 I⁻(aq) ⇄ Mg²⁺(aq) + 2I⁻(aq) + H₂O(l) + SO₂(g)

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Answer: iron (oxides of iron)

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3 years ago
combustion analysis of a hydrocarbon produced 33.01g CO2 and 13.51g H2O. Calculate the empirical formula for the hydrocarbon
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Answer:

\rm CH_2.

Explanation:

Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be \rm CO_2 and \rm H_2O. The \rm C and \rm H would come from the hydrocarbon, while the \rm O atoms would come from oxygen.

Look up the relative atomic mass of these three elements on a modern periodic table:

  • \rm C: 12.011.
  • \rm H: 1.008.
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Calculate the molar mass of \rm CO_2 and \rm H_2O:

M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}.

M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}

Calculate the number of moles of \rm CO_2 molecules in 33.01\; \rm g of \rm CO_2\!:

\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol.

Similarly, calculate the number of moles of \rm H_2O molecules in 13.51\; \rm g of \rm H_2O\!:

\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol.

Note that there is one carbon atom in every \rm CO_2 molecule. Approximately0.7501\; \rm mol of \rm CO_2\! molecules would correspond to the same number of \rm C atoms. That is: n(\mathrm{C}) \approx 0.7501\; \rm mol.

On the other hand, there are two hydrogen atoms in every \rm H_2O molecule. approximately 0.7499\; \rm mol of \rm H_2O molecules would correspond to twice as many \rm H\! atoms. That is: n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol.

The ratio between the two is: n(\mathrm{C}): n(\mathrm{H}) \approx 1:2.

The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be \rm CH_2.

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