Answer:
MgCO₃
Explanation:
From the question given above, we obtained:
MgF₂ + Li₂CO₃ —> __ + 2LiF
The missing part of the equation can be obtained by writing the ionic equation for the reaction between MgF₂ and Li₂CO₃. This is illustrated below:
MgF₂ (aq) —> Mg²⁺ + 2F¯
Li₂CO₃ (aq) —> 2Li⁺ + CO₃²¯
MgF₂ + Li₂CO₃ —>
Mg²⁺ + 2F¯ + 2Li⁺ + CO₃²¯ —> Mg²⁺CO₃²¯ + 2Li⁺F¯
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Now, we share compare the above equation with the one given in the question above to obtain the missing part. This is illustrated below:
MgF₂ + Li₂CO₃ —> __ + 2LiF
MgF₂ + Li₂CO₃ —> MgCO₃ + 2LiF
Therefore, the missing part of the equation is MgCO₃
In chemistry, the molar mass M is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by its amount of substance. The base SI unit for molar mass is kg/mol. However, for historical reasons, molar masses are almost always expressed in g/mol.
Hope this helped!
Good luck :p
~Emmy <3
Answer:
The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.
Explanation:
You know the balanced reaction:
4 NA + O₂ ⟶ 2 Na₂O
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction) react and are produced:
- Na: 4 moles
- O₂: 1 mole
- Na₂O: 2 moles
Being:
the molar mass of the compounds participating in the reaction is:
- Na: 23 g/mole
- O₂: 2*16 g/mole= 32 g/mole
- Na₂O: 2*23 g/mole +16 g/mole= 62 g/mole
Then by stoichiometry of the reaction they react and are produced:
- Na: 4 moles* 23 g/mole= 92 g
- O₂: 1 mole*32 g/mole= 32 g
- Na₂O: 2 moles* 62 g/mole= 124 g
Then you can apply the following rule of three: if 92 grams of Na produce 124 grams of Na₂O, 4 grams of Na, how much mass of Na₂O does it produce?

mass of Na₂O=5.39 g
<em><u>The mass of Na₂O that can be produced by the chemical reaction of 4.0 grams of sodium with excess oxygen in the reaction is 5.39 grams.</u></em>