A 5.000 g sample of a mixture which contains MgCO3 and sand (SiO2) was heated for 2.30 hours until no further reaction occurred. It was then cooled and the residue, which contained MgO and unchanged sand, was weighed. The reaction is: MgCO3(s)  MgO(s) + CO2(g). The residue remaining (minus the CO2 which was completely driven off) weighed 3.397 grams. Calculate the percent, by weight, of MgCO3 in the original sample.

Question

A 5.000 g sample of a mixture which contains MgCO3 and sand (SiO2) was heated for 2.30 hours until no further reaction occurred. It was then cooled and the residue, which contained MgO and unchanged sand, was weighed. The reaction is: MgCO3(s)  MgO(s) + CO2(g). The residue remaining (minus the CO2 which was completely driven off) weighed 3.397 grams. Calculate the percent, by weight, of MgCO3 in the original sample.

Answer 1

MgCO3 + SiO2 + heat ==> MgO + CO2 + SiO2

Loss in mass = 5.000-3.397 = 1.603 g = mass CO2.
mols CO2 = 1.603/44 = ?
mols MgCO3 = the same (from the equation).
mass MgCO3 = mols x molar mass
%MgCO3 = (mass MgCO3/mass sample)*100 = ?

Answer 2

THANK YOU SO MUCH!!

Answer 3

To find the percent, by weight, of MgCO3 in the original sample, we can use the information given.

Step 1: Calculate the mass of MgO formed during the reaction.
According to the reaction, 1 mol of MgCO3 reacts to form 1 mol of MgO.
The molar mass of MgCO3 = 24.31 g/mol (Mg) + 12.01 g/mol (C) + 3(16.00 g/mol) (O) = 84.31 g/mol
The molar mass of MgO = 24.31 g/mol (Mg) + 16.00 g/mol (O) = 40.31 g/mol

From the reaction, we know that the molar ratio of MgCO3 to MgO is 1:1.
Therefore, the mass of MgO formed = mass of MgCO3 reacted = (5.000 g - 3.397 g) = 1.603 g

Step 2: Calculate the moles of MgCO3 reacted.
Moles of MgCO3 = mass of MgCO3 / molar mass of MgCO3
Moles of MgCO3 = 1.603 g / 84.31 g/mol = 0.019 moles

Step 3: Calculate the percent, by weight, of MgCO3 in the original sample.
Percent of MgCO3 = (mass of MgCO3 / mass of original sample) x 100
Percent of MgCO3 = (0.019 moles x 84.31 g/mol) / 5.000 g x 100
Percent of MgCO3 = 31.988%

Therefore, the percent, by weight, of MgCO3 in the original sample is approximately 31.99%.

Answer 4

To calculate the percent by weight of MgCO3 in the original sample, you need to determine the amount of MgCO3 present in the mixture and divide it by the initial weight of the sample.

Given:
- Initial weight of the sample = 5.000 g
- Weight of the residue remaining (minus the CO2) = 3.397 g

First, we need to find the weight of MgCO3 that reacted and formed MgO:
Weight of CO2 released = Initial weight of the sample - Weight of residue remaining
= 5.000 g - 3.397 g
= 1.603 g

Since the balanced equation for the reaction is:
MgCO3(s) -> MgO(s) + CO2(g)
The molar ratio between MgCO3 and CO2 is 1:1.

Next, we need to convert the weight of CO2 to moles. To do that, we need to know the molar mass of CO2, which is 44.01 g/mol.

Moles of CO2 = Weight of CO2 / Molar mass of CO2
= 1.603 g / 44.01 g/mol
= 0.0364 mol

Since the molar ratio between MgCO3 and CO2 is 1:1, the moles of MgCO3 reacted are also 0.0364 mol.

Now, we can calculate the molar mass of MgCO3. Magnesium (Mg) has a molar mass of 24.31 g/mol, carbon (C) has a molar mass of 12.01 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol.

Molar mass of MgCO3 = Molar mass of Mg + Molar mass of C + 3 * Molar mass of O
= 24.31 g/mol + 12.01 g/mol + 3 * 16.00 g/mol
= 84.31 g/mol

Finally, we can calculate the weight of MgCO3 in the original sample:
Weight of MgCO3 = Moles of MgCO3 * Molar mass of MgCO3
= 0.0364 mol * 84.31 g/mol
= 3.063 g

Now we can calculate the percent by weight of MgCO3 in the original sample:
Percent by weight = (Weight of MgCO3 / Initial weight of the sample) * 100
= (3.063 g / 5.000 g) * 100
= 61.26%

Therefore, the percent by weight of MgCO3 in the original sample is approximately 61.26%.