Chemistry, 26.10.2019 00:43 mckadams02

Dolomite is a mixed carbonate of calcium and magnesium that decomposes to co2 and the metal oxides mgo and cao upon heating. when 11.15 g of dolomite is heated, 5.70 g of mgo and cao are produced. what is percent by mass of mgco3 in the original sample of dolomite?

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Answer from: luusperezzz

The percent by mass of MgCO_3 in dolomite is 59.5 %

Explanation:

We are given:

Mass of MgO and CaO produced = 5.70 g

Mass of dolomite = 11.15 g

We know that:

Molar mass of magnesium carbonate = 84.314 g/mol

Molar mass of calcium carbonate = 100.087 g/mol

Molar mass of magnesium oxide = 40.304 g/mol

Molar mass of calcium oxide = 56.077 g/mol

Let the mass of magnesium carbonate in given sample of dolomite be 'x' grams and that of calcium carbonate be '(11.15 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of magnesium carbonate = $\frac{x}{84.314}$ moles

Moles of calcium carbonate = $\frac{(11.15-x)}{100.087}$ moles

The chemical equation for the heating of magnesium carbonate follows the equation:

$MgCO_3\rightarrow MgO+CO_2$

By Stoichiometry of the reaction:

1 mole of magnesium carbonate produces 1 mole of magnesium oxide

So, $\frac{x}{84.314}$ moles of magnesium carbonate will produce = $\frac{1}{1}\times \frac{x}{84.314}=\frac{x}{84.314}mol$ of magnesium oxide

Mass of magnesium oxide = $\text{Moles of magnesium oxide}\times \text{Molar mass of magnesium oxide}$

Mass of magnesium oxide = $(\frac{x}{84.314}\times 40.304)g$

The chemical equation for the heating of calcium carbonate follows the equation:

$CaCO_3\rightarrow CaO+CO_2$

By Stoichiometry of the reaction:

1 mole of calcium carbonate produces 1 mole of calcium oxide

So, $\frac{(11.15-x)}{100.087}$ moles of calcium carbonate will produce = $\frac{1}{1}\times \frac{(11.15-x)}{100.087}=\frac{(11.15-x)}{100.087}mol$ of calcium oxide

Mass of calcium oxide = $\text{Moles of calcium oxide}\times \text{Molar mass of calcium oxide}$

Mass of calcium oxide = $(\frac{(11.15-x)}{100.087}\times 56.077)g$

We are given:

Mass of MgO and CaO produced = 5.70 g

$(\frac{x}{84.314}\times 40.304)+(\frac{(11.15-x)}{100.087}\times 56.077)=5.70\\\\0.478x+6.244-0.560x=5.70\\\\0.082x=0.544\\\\x=\frac{0.544}{0.082}=6.63g$

To calculate the mass percentage of magnesium carbonate in dolomite, we use the equation:

$\text{Mass percent of magnesium carbonate}=\frac{\text{Mass of magnesium carbonate}}{\text{Mass of dolomite}}\times 100$

Mass of dolomite = 11.15 g

Mass of magnesium carbonate = 6.63 g

Putting values in above equation, we get:

$\text{Mass percent of magnesium carbonate}=\frac{6.63g}{11.15g}\times 100=59.5\%$

Hence, the mass percent of magnesium carbonate in given amount of dolomite is 59.5 %

Answer from: Quest

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%

Answer from: Quest

30litre of li2o is 1.0002mole approximately 1 mole

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