what volume of 2.50 M HCl in liters is needed to react completly(with noting left over) with .750 L of 0.500MNacCO3
and the equation is
2HCL+Na2CO3=2NaCl+H2O+CO2
To find the volume of 2.50 M HCl needed to react completely with 0.750 L of 0.500 M Na2CO3, you can follow these steps:
Step 1: Write the balanced chemical equation for the reaction between HCl and Na2CO3:
2 HCl + Na2CO3 → 2 NaCl + H2O + CO2
Step 2: Determine the stoichiometric ratio between HCl and Na2CO3 from the balanced equation. In this case, for every 2 moles of HCl, we need 1 mole of Na2CO3.
Step 3: Convert the given volumes of the solutions into moles using the formula:
moles = concentration (M) × volume (L)
For HCl:
moles of HCl = 2.50 M × volume of HCl (L)
For Na2CO3:
moles of Na2CO3 = 0.500 M × 0.750 L
Step 4: Use the stoichiometric ratio to determine the volume of HCl needed. Since we need 2 moles of HCl for every 1 mole of Na2CO3, we can set up the following ratio:
(2.50 M × volume of HCl) / (2 moles HCl) = (0.500 M × 0.750 L) / (1 mole Na2CO3)
Step 5: Rearrange the equation to solve for the volume of HCl:
volume of HCl = (0.500 M × 0.750 L × 2 moles HCl) / (2.50 M × 1 mole Na2CO3)
Step 6: Calculate the volume of HCl:
volume of HCl = (0.500 M × 0.750 L × 2) / (2.50 M)
The units M (molarity) cancel out, leaving the final answer in liters.
Simplifying the equation:
volume of HCl = (0.500 × 0.750 × 2) / 2.50
By calculating the expression above:
volume of HCl = 0.750 L
Therefore, 0.750 liters of 2.50 M HCl is needed to react completely with 0.750 L of 0.500 M Na2CO3.