Calculate the number of moles of mgcl2 is produced by reacting 168g ofmgco3 with excess hcl(mg=2,c=12,o=16,cl=35.5)
balance the reaction
MgCO3+2HCl>>>MgCl2+ CO2+H2O
moles of MgCO3=168/84.3=2.0
then from the balance equation should get 2 moles os MgCl2
To calculate the number of moles of MgCl2 produced, we need to follow these steps:
Step 1: Determine the molar mass of MgCO3.
The molar mass of MgCO3 can be calculated by adding the atomic masses of its constituent elements:
MgCO3 = (1 × Mg) + (1 × C) + (3 × O)
The atomic mass of each element is:
Mg = 24.31 g/mol
C = 12.01 g/mol
O = 16.00 g/mol
Substituting these values into the formula, we get:
Molar mass of MgCO3 = (1 × 24.31) + (1 × 12.01) + (3 × 16.00) g/mol
Step 2: Calculate the number of moles of MgCO3.
We can use the formula:
Number of moles = Mass / Molar mass
Mass of MgCO3 = 168 g (given)
Molar mass of MgCO3 = Calculated in step 1
Substituting the values into the formula, we get:
Number of moles of MgCO3 = 168 g / Molar mass of MgCO3
Step 3: Determine the balanced chemical equation for the reaction between MgCO3 and HCl.
The balanced chemical equation for the reaction is:
MgCO3 + 2HCl → MgCl2 + CO2 + H2O
From the balanced equation, we can see that 1 mole of MgCO3 produces 1 mole of MgCl2.
Step 4: Finally, calculate the number of moles of MgCl2 produced.
Since we have calculated the number of moles of MgCO3 in step 2, we can conclude that the number of moles of MgCl2 produced will be the same. So, the number of moles of MgCl2 produced is equal to the number of moles of MgCO3, calculated using the formula in step 2.
Note: Excess HCl indicates that HCl is present in excess, meaning it won't be limiting the reaction. Hence, all the MgCO3 will be reacting to form MgCl2.