Question

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following equation: MgCO3 --> MgO CO2

Answer 1

`210.7~g~MgCO_3`

Step-by-step explanation:

We have to start with the reaction:

`MgCO_3~->~MgO~+~CO_2`

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:

`(12*1)+(16*2)=44~g/mol`

In other words:
`1~mol~CO_2=~44~g~CO_2`. With this in mind, we can calculate the moles:

`110~g~CO_2(1~mol~CO_2)/(44~g~CO_2)=25~mol~CO_2`

Now, the molar ratio between carbon dioxide and magnesium carbonate is 1:1, so:

`2.5~mol~CO_2=2.5~mol~MgCO_3`

With the molar mass of
`MgCO_3` (
`(23.3*1)+(12*1)+(16*3)=84.3~g/mol`. With this in mind, we can calculate the grams of magnesium carbonate:

`2.5~mol~MgCO_3(84.3~g~MgCO_3)/(1~mol~MgCO_3)=210.7~g~MgCO_3`

I hope it helps!