Question

what volume of 2.25 m hcl in liters is needed to react completely (with nothing left over) with 0.250 l of 0.500 m na2co3 ?

Answer 1

To react completely with 0.250 L of 0.500 M Na2CO3, you would need approximately 55.6 mL of 2.25 M HCl.

Step-by-step explanation:

To determine the volume of 2.25 M HCl needed to react completely with 0.250 L of 0.500 M Na2CO3, we need to use the balanced chemical equation:

2 HCl(aq) + Na2CO3(aq) → 2 NaCl(aq) + H2O(l) + CO2(g)

From the balanced equation, we can see that the ratio of moles of HCl to Na2CO3 is 2:1. Therefore, the number of moles of HCl needed is half the number of moles of Na2CO3.

We can calculate the number of moles of Na2CO3:

0.500 M Na2CO3 * 0.250 L = 0.125 moles of Na2CO3

Since the ratio of moles of HCl to Na2CO3 is 2:1, we need 0.125 moles of HCl. Using the molarity equation, we can calculate the volume of HCl needed:

0.125 moles HCl / 2.25 M HCl = 0.0556 L or 55.6 mL

Answer 2

To react completely with 0.250 L of 0.500 M Na2CO3, 111 mL of 2.25 M HCl is needed, based on the stoichiometry of the balanced chemical equation.

Step-by-step explanation:

To calculate the volume of 2.25 M HCl needed to react with 0.250 L of 0.500 M Na2CO3, we must first determine the number of moles of Na2CO3 that will be reacting. The balanced chemical equation for the reaction between HCl and Na2CO3 is:

2 HCl(aq) + Na2CO3(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)

From the balanced equation, we can see that it takes 2 moles of HCl to react with 1 mole of Na2CO3. Therefore, we need to calculate the moles of Na2CO3 first:

Moles of Na2CO3 = Molarity × Volume = 0.500 mol/L × 0.250 L = 0.125 moles

Now, to get the moles of HCl needed:

Moles of HCl needed = 0.125 moles Na2CO3 × (2 moles HCl/1 mole Na2CO3) = 0.250 moles

Finally, to find the volume of HCl needed:

Volume of HCl = Moles of HCl / Molarity of HCl = 0.250 moles / 2.25 mol/L

= 0.111 L or 111 mL

Therefore, to completely react with 0.250 L of 0.500 M Na2CO3, you would need 111 mL of 2.25 M HCl.