What volume of 1.25M HCl in liters is needed to react completely (with nothing left over) with 0.750L of 0.300M 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 the defintion of M HCl = mols HCl/L HCl. You know M and mols, solve for L.
To determine the volume of 1.25M HCl needed to react completely with 0.750L of 0.300M Na2CO3, we can use the concept of stoichiometry.
The balanced chemical equation for the reaction between HCl and Na2CO3 is:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
From the equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3. Therefore, the stoichiometric ratio between HCl and Na2CO3 is 2:1.
First, we need to determine the number of moles of Na2CO3 in 0.750L of 0.300M Na2CO3 solution. To do this, we can use the formula:
moles = concentration (M) × volume (L)
moles of Na2CO3 = 0.300M × 0.750L
moles of Na2CO3 = 0.225 moles
Since the stoichiometric ratio between HCl and Na2CO3 is 2:1, we know that we need twice as many moles of HCl as Na2CO3. Therefore, we need:
moles of HCl = 2 × moles of Na2CO3
moles of HCl = 2 × 0.225 moles
moles of HCl = 0.450 moles
Now, let's determine the volume of 1.25M HCl solution needed to have 0.450 moles of HCl:
moles = concentration (M) × volume (L)
0.450 moles = 1.25M × volume (L)
Solving for volume (L):
volume (L) = 0.450 moles / 1.25M
volume (L) = 0.36 L
Therefore, to react completely with 0.750L of 0.300M Na2CO3, we would need approximately 0.36 liters of 1.25M HCl.
To find the volume of a solution needed for a complete reaction, we can use the principle of stoichiometry. First, let's write the balanced chemical equation for the reaction between HCl and Na2CO3:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of Na2CO3 to produce 2 moles of NaCl, 1 mole of H2O, and 1 mole of CO2.
Now let's calculate the number of moles of Na2CO3 present in 0.750 liters of a 0.300M solution:
Molarity (M) = moles of solute / volume of solution (in liters)
0.300M = moles of Na2CO3 / 0.750L
moles of Na2CO3 = 0.300M x 0.750L = 0.225 moles
Since the stoichiometric ratio tells us that 2 moles of HCl react with 1 mole of Na2CO3, we need twice the number of moles of HCl to react completely:
moles of HCl needed = 2 x 0.225 moles = 0.450 moles
Now, we can calculate the volume of 1.25M HCl solution needed to contain 0.450 moles of HCl:
Molarity (M) = moles of solute / volume of solution (in liters)
1.25M = 0.450 moles / volume of HCl solution (in liters)
volume of HCl solution (in liters) = 0.450 moles / 1.25M = 0.36 L
Therefore, a volume of 0.36 liters (or 360 mL) of 1.25M HCl solution is needed to completely react with 0.750L of 0.300M Na2CO3 solution.