Answer:
[tex]210.7~g~MgCO_3[/tex]
Explanation:
We have to start with the reaction:
[tex]MgCO_3~->~MgO~+~CO_2[/tex]
We have the same amount of atoms on both sides, so, we can continue. The next step is to find the number of moles that we have in the 110.0 g of carbon dioxide, to this, we have to know the atomic mass of each atom:
C: 12 g/mol
O: 16 g/mol
Mg: 23.3 g/mol
If we take into account the number of atoms in the formula, we can calculate the molar mass of carbon dioxide:
[tex](12*1)+(16*2)=44~g/mol[/tex]
In other words: [tex]1~mol~CO_2=~44~g~CO_2[/tex]. With this in mind, we can calculate the moles:
[tex]110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2[/tex]
Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:
[tex]2.5~mol~CO_2=2.5~mol~MgCO_3[/tex]
With the molar mass of [tex]MgCO_3[/tex] ([tex](23.3*1)+(12*1)+(16*3)=84.3~g/mol[/tex]. With this in mind, we can calculate the grams of magnesium carbonate:
[tex]2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3[/tex]
I hope it helps!
