What volume of 3.00M HCl in liters is needed to react completely (with nothing left over) with 0.750L of 0.400M Na2CO3?
Na2CO3 + 2HCl ==> 2NaCl + H2O + CO2
mols Na2CO3 = M x L = ?
Using the coefficients in the balanced equation, convert mols Na2CO3 to mols HCl.
Then M HCl = mols HCl/L HCl. YOu know M HCl and mols HCl, solve for L HCl.
To find the volume of the 3.00M HCl required to react completely with 0.750L of 0.400M Na2CO3, we can use the concept of stoichiometry.
Step 1: Write a balanced equation for the reaction.
Na2CO3 + 2HCl -> 2NaCl + H2O + CO2
From the equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.
Step 2: Determine the number of moles of Na2CO3.
To find the moles of Na2CO3, we can use the formula:
moles = concentration × volume
moles of Na2CO3 = 0.400M × 0.750L = 0.300 moles
Step 3: Determine the volume of HCl required.
Since the stoichiometric ratio between Na2CO3 and HCl is 1:2, we can calculate the number of moles of HCl required using the ratio:
moles of HCl = 2 × moles of Na2CO3 = 2 × 0.300 moles = 0.600 moles
Step 4: Determine the volume of 3.00M HCl.
To find the volume of 3.00M HCl, we can use the formula:
volume = moles / concentration
volume of 3.00M HCl = 0.600 moles / 3.00M = 0.200 L
Therefore, the volume of 3.00M HCl required to react completely with 0.750L of 0.400M Na2CO3 is 0.200 liters.