magnesium trioxocarbonate (IV) decomposed according to the equation MgCO3(s) -MgO2(s) + CO2(g).if 6.0g of magnesium trioxocarbonate (IV) is completely decomposed ,calculate :i. the mass of carbon (IV) oxide produced
To calculate the mass of carbon (IV) oxide produced when 6.0g of magnesium trioxocarbonate (IV) decomposes, we need to determine the molar ratio between magnesium trioxocarbonate (IV) and carbon (IV) oxide.
The balanced chemical equation for the decomposition of magnesium trioxocarbonate (IV) is:
MgCO3(s) → MgO(s) + CO2(g)
From the equation, we can see that the molar ratio of magnesium trioxocarbonate (IV) to carbon (IV) oxide is 1:1. This means that for every 1 mole of magnesium trioxocarbonate (IV) that decomposes, 1 mole of carbon (IV) oxide is produced.
To calculate the moles of magnesium trioxocarbonate (IV) in 6.0g, we need to divide the mass by the molar mass of magnesium trioxocarbonate (IV).
The molar mass of magnesium trioxocarbonate (IV) can be calculated by adding the molar masses of its constituent elements:
MgCO3 = (24.31 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol) = 84.31 g/mol
Now, let's calculate the moles of magnesium trioxocarbonate (IV):
Moles of MgCO3 = Mass / Molar mass
= 6.0g / 84.31 g/mol
≈ 0.071 moles
Since the molar ratio of MgCO3 to CO2 is 1:1, the moles of carbon (IV) oxide produced will also be 0.071 moles.
To calculate the mass of carbon (IV) oxide produced, we can use the molar mass of carbon (IV) oxide, which is 44.01 g/mol.
Mass of CO2 = Moles of CO2 * Molar mass
= 0.071 moles * 44.01 g/mol
≈ 3.13 g
Therefore, the mass of carbon (IV) oxide produced when 6.0g of magnesium trioxocarbonate (IV) decomposes is approximately 3.13g.
To calculate the mass of carbon (IV) oxide (CO2) produced when 6.0g of magnesium trioxocarbonate (IV) decomposes, we need to determine the molar ratio of magnesium trioxocarbonate (IV) to carbon (IV) oxide.
First, let's calculate the molar mass of magnesium trioxocarbonate (MgCO3):
Molar mass of Mg = 24.31 g/mol
Molar mass of C = 12.01 g/mol
3 x Molar mass of O = 3 x 16.00 g/mol = 48.00 g/mol
Adding them up:
Molar mass of MgCO3 = 24.31 g/mol + 12.01 g/mol + 48.00 g/mol = 84.32 g/mol
Next, we can calculate the molar ratio between MgCO3 and CO2:
From the balanced equation: MgCO3(s) → MgO(s) + CO2(g)
The coefficient in front of MgCO3 is 1, and the coefficient in front of CO2 is also 1, indicating a 1:1 molar ratio. This means that for every 1 mole of MgCO3 decomposed, 1 mole of CO2 is produced.
To find the number of moles of MgCO3, divide the given mass by the molar mass:
Number of moles of MgCO3 = 6.0 g / 84.32 g/mol = 0.071 moles
Since the molar ratio is 1:1, the number of moles of CO2 produced is also 0.071 moles.
Now, to find the mass of CO2 produced, multiply the number of moles by the molar mass of CO2:
Mass of CO2 = 0.071 moles × 44.01 g/mol (molar mass of CO2) = 3.12 g
Therefore, the mass of carbon (IV) oxide produced when 6.0g of magnesium trioxocarbonate (IV) decomposes is 3.12g.