how many grams of HCL are consumed in the reaction of 378 grams of a mixture containing 18.2 % MgCO3 and 81.1% MgOH2 by mass
To determine the amount of HCl consumed in the reaction, we need to use stoichiometry and the balanced chemical equation of the reaction. However, you haven't provided the chemical equation of the reaction between HCl and the mixture containing MgCO3 and MgOH2. Therefore, I will assume a balanced equation for the reaction:
MgCO3 + 2HCl -> MgCl2 + CO2 + H2O
Step 1: Calculate the mass of MgCO3 and MgOH2 in the mixture
Given that the mass of the mixture is 378 grams and it contains 18.2% MgCO3 and 81.1% MgOH2, we can calculate the masses of these compounds:
Mass of MgCO3 = 378 grams × 0.182 = 68.796 grams
Mass of MgOH2 = 378 grams × 0.811 = 306.358 grams
Step 2: Convert the masses of MgCO3 and MgOH2 to moles
To use stoichiometry, we need to convert the masses of MgCO3 and MgOH2 to moles using their molar masses. The molar masses of MgCO3 and MgOH2 are:
MgCO3 = 24.31 g/mol (Molar mass of Mg) + 12.01 g/mol (Molar mass of C) + 3 × 16.00 g/mol (Molar mass of O) = 84.31 g/mol
MgOH2 = 24.31 g/mol (Molar mass of Mg) + 2 × 16.00 g/mol (Molar mass of O) + 2 × 1.01 g/mol (Molar mass of H) = 58.32 g/mol
Now, we can calculate the number of moles:
Moles of MgCO3 = Mass of MgCO3 / Molar mass of MgCO3 = 68.796 g / 84.31 g/mol ≈ 0.816 moles
Moles of MgOH2 = Mass of MgOH2 / Molar mass of MgOH2 = 306.358 g / 58.32 g/mol ≈ 5.257 moles
Step 3: Determine the stoichiometry of the reaction
The balanced chemical equation shows that 1 mole of MgCO3 reacts with 2 moles of HCl.
Step 4: Calculate the amount of HCl consumed
Using the stoichiometry, we can determine the number of moles of HCl consumed in the reaction:
Moles of HCl = 2 × Moles of MgCO3 = 2 × 0.816 moles ≈ 1.632 moles
Step 5: Convert moles of HCl to grams
To convert moles of HCl to grams, we need to multiply by the molar mass of HCl. The molar mass of HCl is:
HCl = 1.01 g/mol (Molar mass of H) + 35.45 g/mol (Molar mass of Cl) = 36.46 g/mol
Finally, we can calculate the grams of HCl consumed:
Grams of HCl = Moles of HCl × Molar mass of HCl = 1.632 moles × 36.46 g/mol ≈ 59.53 grams
Therefore, approximately 59.53 grams of HCl are consumed in the reaction.