What volume of 2.25\it M \rm HCl in liters is needed to react completely (with nothing left over) with 0.750L of 0.200\it M \rm Na_2CO_3?
To solve this problem, we can use the balanced chemical equation between HCl and Na2CO3:
2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2
Based on the equation, we can see that 2 moles of HCl reacts with 1 mole of Na2CO3.
First, we need to calculate the number of moles of Na2CO3 using its concentration and volume:
moles of Na2CO3 = concentration * volume
= 0.200 M * 0.750 L
= 0.150 moles
Since the stoichiometry between HCl and Na2CO3 is 2:1, we need twice as many moles of HCl as Na2CO3 to completely react.
moles of HCl = 2 * moles of Na2CO3
= 2 * 0.150 moles
= 0.300 moles
Finally, we can use the concentration of HCl to find the volume needed to react completely:
volume of HCl = moles of HCl / concentration
= 0.300 moles / 2.25 M
= 0.133 L
Therefore, a volume of 0.133 liters of 2.25 M HCl is needed to react completely with 0.750 liters of 0.200 M Na2CO3.
To determine the volume of 2.25 M HCl needed to react completely with 0.750 L of 0.200 M Na2CO3, we need to use the balanced chemical equation and stoichiometry.
First, let's write the balanced chemical equation for the reaction between HCl and Na2CO3:
2 HCl + Na2CO3 -> 2 NaCl + H2O + CO2
From the balanced equation, we can see that it takes 2 moles of HCl to react with 1 mole of Na2CO3.
To solve this problem and find the volume of HCl needed, we'll follow these steps:
Step 1: Calculate the number of moles of Na2CO3 in 0.750 L of the 0.200 M solution.
Moles = Volume x Concentration
Moles of Na2CO3 = 0.750 L x 0.200 mol/L
Step 2: Use the stoichiometry of the balanced chemical equation to find the number of moles of HCl needed for the reaction.
Since the stoichiometry is 2:1 (2 moles of HCl react with 1 mole of Na2CO3), we multiply the number of moles of Na2CO3 by 2 to get the moles of HCl.
Moles of HCl = Moles of Na2CO3 x 2
Step 3: Calculate the volume (in liters) of 2.25 M HCl needed using the concentration of HCl and the moles from Step 2.
Volume = Moles / Concentration
Volume of HCl = (Moles of HCl) / 2.25 mol/L
By following these steps, we can find the volume of 2.25 M HCl in liters needed to react completely with 0.750 L of 0.200 M Na2CO3.