Chemistry: I d love some help please thank you

I d love some help please thank you - id10434331, tacojordan4128

Ответ:

kahmir15

27.05.2021

Chemistry

210.7~g~MgCO_3

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:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

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:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

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