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lara31 [8.8K]
3 years ago
10

Write the net ionic equation for this reaction occurring in water: ammonium fluoride and magnesium chloride are mixed to form ma

gnesium fluoride and ammonium chloride. 1. no reaction occurs. 2. 2 nh+ 4 + 2 cl− → 2 nh4cl 3. 2 f− + mg2+ → mgf2 4. 2 nh4f + 2 cl− → 2 f− + 2 nh4cl 5. 2 f− + mgcl2 → mgf2 + 2 cl−
Chemistry
1 answer:
MAXImum [283]3 years ago
6 0

Correct answer: 3. Mg^{2+}(aq) + 2 F^{-}(aq) --> MgF_{2}(s)

The given chemical reaction is between ammonium fluoride and magnesium chloride to form magnesium fluoride and ammonium chloride.

The balanced chemical equation representing the reaction will be,

2NH_{4}F (aq) + MgCl_{2}(aq) -->MgF_{2} (s) + 2NH_{4}Cl (aq)

The complete ionic equation for the reaction: All the compounds soluble in water (aqueous) will split into ions, MgF_{2} will not split into ions as it is insoluble in water.

2 NH_{4}^{+}(aq) + 2 F^{-}(aq) + Mg^{2+}(aq) + 2 Cl^{-}(aq) --> MgF_{2}(s) + 2 NH_{4}^{+}(aq) + 2 Cl^{-}(aq)

The net ionic equation will be:

Mg^{2+}(aq) + 2 F^{-}(aq) --> MgF_{2}(s)

Here the spectator ions are NH_{4}^{+} and Cl^{-}


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How many grams of h2 will be produced if 175g of HCI are allowed to react completely with sodium
Sedbober [7]

Answer:

4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium

Explanation:

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) you can see that the following amounts in moles of each compound react and are produced:

  • HCl: 2 moles
  • Na: 1 mole
  • NaCl: 2 moles
  • H₂: 1 mole

You know the following masses of each element:

  • H: 1 g/mole
  • Cl: 35.45 g/mole
  • Na: 23 g/mole

So, the molar mass of each compound participating in the reaction is:

  • HCl: 1 g/mole + 35.45 g/mole= 36.45 g/mole
  • Na: 23 g/mole
  • NaCl: 23 g/mole + 35.45 g/mole= 58.45 g/mole
  • H₂: 2* 1 g/mole= 2 g/mole

Then, by stoichiometry of the reaction, the following amounts in grams of each of the compounds participating in the reaction react and are produced:

  • HCl: 2 moles* 36.45 g/mole= 72.9 g
  • Na: 1 mole* 23 g/mole= 23 g
  • NaCl: 2 moles* 58.45 g/mole= 116.9 g
  • H₂: 1 mole* 2 g/mole= 2 g

So, a rule of three applies as follows: if by stoichiometry, when reacting 72.9 grams of HCl 2 grams of H₂ are formed, when reacting 175 grams of HCl how much mass of H₂ will be formed?

mass of H_{2} =\frac{175 g of HCl*2g ofH_{2} }{72.9 g of HCl}

mass of H₂= 4.8 g

<u><em>4.8 grams of H₂ will be produced if 175g of HCI are allowed to react completely with sodium</em></u>

3 0
3 years ago
Read 2 more answers
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1) Finding
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Molarity= \frac{0.178 mol}{L} , &#10;&#10;or 0.178 M&#10;


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A sample of oxygen gas occupies a volume of 250 mL at .8 atm of pressure. What volume will it occupy at 1.2 atm of pressure?
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Answer:

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