Sodium carbonate reacts with a solution of HCl solution with the evolution of gas (CO2) to yield a solution that is neutral. If 0.265 grams of sodium carbonate are required to neutralize 1.00 mL of HCl solution, what mass (in grams) of sodium carbonate will be required to nutralize 1.00 x 106 µL (microliters) of the same HCl solution? Remember to use the balanced equation for this reaction. DO NOT use units, exponential notation or commas in your answer; NUMBERS ONLY.

Question

Sodium carbonate reacts with a solution of HCl solution with the evolution of gas (CO2) to yield a solution that is neutral. If 0.265 grams of sodium carbonate are required to neutralize 1.00 mL of HCl solution, what mass (in grams) of sodium carbonate will be required to nutralize 1.00 x 106 µL (microliters) of the same HCl solution? Remember to use the balanced equation for this reaction. DO NOT use units, exponential notation or commas in your answer; NUMBERS ONLY.

Answer 1

do it as a proportion:

x/.265g = 1E-6L/1E-3L

x= .265 mg
x=.000265grams
check my thinking.

Answer 2

To find the mass of sodium carbonate required to neutralize 1.00 x 10⁶ µL of the HCl solution, we need to first determine the molar ratio between sodium carbonate (Na₂CO₃) and HCl using the balanced equation for the reaction.

The balanced equation for the reaction between sodium carbonate and HCl is:

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

From the equation, we can see that 1 mole of sodium carbonate reacts with 2 moles of HCl.

Now, let's calculate the moles of HCl that are neutralized by 0.265 grams of sodium carbonate:

First, we need to find the molar mass of sodium carbonate (Na₂CO₃):
Molar mass of Na = 22.99 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol

Molar mass of Na₂CO₃ = (2 * 22.99) + 12.01 + (3 * 16.00) = 105.99 g/mol

Now, let's calculate the moles of Na₂CO₃ using the given mass:
moles of Na₂CO₃ = mass / molar mass
moles of Na₂CO₃ = 0.265 g / 105.99 g/mol

Next, using the molar ratio from the balanced equation, we can calculate the moles of HCl neutralized by the given moles of Na₂CO₃:

moles of HCl = 2 * moles of Na₂CO₃

Now, let's calculate the moles of HCl neutralized by 0.265 g of Na₂CO₃.

Finally, we can use the moles of HCl calculated above to determine the mass of Na₂CO₃ required to neutralize 1.00 x 10⁶ µL of the same HCl solution:

mass of Na₂CO₃ = moles of HCl * molar mass of Na₂CO₃

Remember to convert µL to mL by dividing by 1,000 since 1 mL = 1,000 µL.

Now, you can complete the calculations using the given values and the steps explained above.