If 950 mL of 0.800 M aqueous Na2CO3 and 920 mL of 0.565 M aqueous HCl are reacted stoichiometrically according to the balanced equation, how many milliliters of 0.800 M aqueous Na2CO3 remain? Round your answer to 3 significant figures.
Na2CO3(aq) + 2HCl(aq) ¡æ CO2(g) + 2NaCl(aq) + H2O(l)
Molar Mass (g/mol)
Na2CO3 105.99
HCl 36.461
Molar Volume (L)
22.4 at STP
Gas Constant
(L.atm.mol-1.K-1)
0.0821
To solve this problem, we can use the concept of stoichiometry and the given concentrations of Na2CO3 and HCl to determine the amount of each substance present before and after the reaction.
Step 1: Calculate the initial moles of Na2CO3 and HCl.
Number of moles = concentration × volume (in liters)
Moles of Na2CO3 = 0.800 M × 0.950 L = 0.760 mol
Moles of HCl = 0.565 M × 0.920 L = 0.520 mol
Step 2: Determine the limiting reactant.
To find the limiting reactant, we compare the moles of each reactant with the stoichiometric ratio provided by the balanced equation. From the equation, we can see that 1 mole of Na2CO3 reacts with 2 moles of HCl.
Since the stoichiometric ratio is 1:2, we can convert the moles of Na2CO3 to moles of HCl:
0.760 mol Na2CO3 × (2 mol HCl / 1 mol Na2CO3) = 1.52 mol HCl
Since we have less than 1.52 mol of HCl, Na2CO3 is the limiting reactant.
Step 3: Calculate the moles of Na2CO3 that reacted.
From the stoichiometry, we know that 1 mole of Na2CO3 reacts with 1 mole of CO2 and 1 mole of Na2CO3 produces 2 moles of NaCl.
Thus, the moles of Na2CO3 that reacted is equal to the moles of CO2 produced, which is also equal to half the moles of HCl used.
Moles of Na2CO3 reacted = 0.520 mol HCl / 2 = 0.260 mol
Step 4: Calculate the remaining moles of Na2CO3.
Moles of Na2CO3 remaining = initial moles of Na2CO3 - moles of Na2CO3 reacted
Moles of Na2CO3 remaining = 0.760 mol - 0.260 mol = 0.500 mol
Step 5: Calculate the volume of Na2CO3 remaining.
Volume of Na2CO3 remaining = moles of Na2CO3 remaining / concentration of Na2CO3
Volume of Na2CO3 remaining = 0.500 mol / 0.800 M = 0.625 L
Step 6: Convert the volume to milliliters.
Volume of Na2CO3 remaining = 0.625 L × 1000 mL/L = 625 mL
Therefore, 625 mL of 0.800 M aqueous Na2CO3 remain after the reaction.
To solve this problem, we need to use the concept of stoichiometry. Stoichiometry is a branch of chemistry that deals with the relationship between the quantities of reactants and products in a chemical reaction.
First, let's determine the limiting reactant. The limiting reactant is the reactant that is completely consumed in the reaction and determines the maximum amount of product that can be formed. In this case, we have 950 mL of 0.800 M Na2CO3 and 920 mL of 0.565 M HCl.
To find the limiting reactant, we need to calculate the number of moles of each reactant. We can use the formula:
moles = concentration (M) x volume (L)
For Na2CO3:
moles of Na2CO3 = 0.800 M x 0.950 L = 0.760 mol
For HCl:
moles of HCl = 0.565 M x 0.920 L = 0.5198 mol
According to the balanced equation, the stoichiometric ratio between Na2CO3 and HCl is 1:2. This means that for every 1 mole of Na2CO3, we need 2 moles of HCl.
Since the stoichiometric ratio is 1:2, we compare the number of moles of Na2CO3 and HCl. The mole ratio is 0.760 mol of Na2CO3 to 0.5198 mol of HCl. Since we need twice as many moles of HCl, we can say that HCl is the limiting reactant.
Now, we can calculate the amount of Na2CO3 that reacts with the limiting reactant (HCl). From the balanced equation, we know that the mole ratio of Na2CO3 to HCl is 1:2.
moles of Na2CO3 reacted = 0.5198 mol of HCl x (1 mol of Na2CO3 / 2 mol of HCl) = 0.2599 mol of Na2CO3
Now, let's calculate the volume of remaining Na2CO3 solution. We have 950 mL (0.950 L) of 0.800 M Na2CO3 initially, and we know that 0.2599 moles of Na2CO3 reacted.
moles of Na2CO3 remaining = 0.760 mol - 0.2599 mol = 0.5001 mol
To calculate the volume, we can use the formula:
volume (L) = moles / concentration (M)
volume = 0.5001 mol / 0.800 M = 0.6251 L
Finally, we convert the volume from liters to milliliters:
volume = 0.6251 L x 1000 mL/L = 625.1 mL
Rounding to three significant figures, the answer is 625 mL.
Therefore, after the reaction, 625 milliliters of the 0.800 M aqueous Na2CO3 solution remain.