Question

1 3. MgCO3co → Mg Ocss + CO2 caj a.) To prepare 70.0g start with how many of g Mgo you of Mg CO3 ? must b. How many as a by g of CO2 would product? be generated

Answer 1

Ans 3

Part a

The given reaction

MgCO3 = MgO + CO2

Mass of MgO = 70 g

Moles of MgO = mass/molecular weight

= (70g) / (40.3044 g/mol)

= 1.737 mol

From the stoichiometry of the reaction

1 mol MgO formed from = 1 mol MgCO3

Moles of MgCO3 reacted = 1.737 mol

Mass of MgCO3 = moles x molecular weight

= 1.737 mol x 84.3139 g/mol

= 146.43 g

Part b

Moles of CO2 produced = moles of MgO formed = 1.737 mol

Mass of CO2 produced = moles x molecular weight

= 1.737 mol x 44 g/mol

= 76.42 g