If 10.0 grams of sodium bicarbonate reacts with 50.0mL of 2M acetic acid, what is the limiting reactant? what reactant is in excess, and by how much? What volume of carbon dioxide gas would form?

Write the equation and balance it. Then work TWO stoichiometry problems, one each with NaHCO3 to produce CO2 and the other with acetic acid to produce CO2. The two answers likely will not be the same; the correct answer in limiting reagent problems is ALWAYS the smaller number and the reagent producing that value is the limiting reagent. Follow the steps in the link.

http://www.jiskha.com/science/chemistry/stoichiometry.html

To determine the limiting reactant, we need to compare the number of moles of each reactant and see which one will be completely consumed in the reaction. The balanced chemical equation for the reaction between sodium bicarbonate (NaHCO3) and acetic acid (CH3COOH) is:

NaHCO3 + CH3COOH -> CH3COONa + H2O + CO2

First, let's calculate the number of moles of each reactant.

1) Sodium Bicarbonate (NaHCO3):
The molar mass of NaHCO3 is 84.01 g/mol.
Given that the mass of NaHCO3 is 10.0 grams, we can calculate the moles using the formula:

moles = mass (g) / molar mass (g/mol)

moles of NaHCO3 = 10.0 g / 84.01 g/mol

2) Acetic Acid (CH3COOH):
The concentration of acetic acid is given as 2M (2 moles per liter or 2 mol/L). Since the volume is given in mL, we need to convert it to liters:

volume (L) = 50.0 mL / 1000 mL/L

Now we can calculate the number of moles of acetic acid using the formula:

moles = concentration (mol/L) * volume (L)

moles of CH3COOH = 2 mol/L * 50.0 mL / 1000 mL/L

Now that we know the moles of each reactant, we can compare them to determine the limiting reactant.

Next, we need to compare the moles of each reactant using the mole ratio from the balanced chemical equation. The balanced equation tells us that the stoichiometric ratio between NaHCO3 and CH3COOH is 1:1. This means one mole of NaHCO3 will react with one mole of CH3COOH.

If the moles of NaHCO3 are equal to or greater than the moles of CH3COOH, then NaHCO3 is not the limiting reactant and vice versa.

Now that we've determined the limiting reactant, we can calculate the excess reactant and the amount of CO2 gas produced.

To determine the excess reactant, subtract the moles of the limiting reactant from the moles of the excess reactant. This will give us the moles of the excess reactant.

To calculate the volume of carbon dioxide (CO2) gas formed, we can use the ideal gas law. However, we need to know the temperature and pressure conditions of the reaction to do so.