In the laboratory, acid spills are often neutralized by adding sodium bicarbonate. What mass of bicarbonate reacts with 225 mL of 6.00 M HCl?
Here is a sample stoichiometry problem I've posted. Just follow the steps. Remember moles = M x L.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the mass of sodium bicarbonate (NaHCO3) that reacts with hydrochloric acid (HCl), you need to set up a balanced chemical equation and use stoichiometry.
The balanced equation for the reaction between HCl and NaHCO3 is:
HCl + NaHCO3 -> NaCl + CO2 + H2O
From the balanced equation, you can see that for every 1 mole of HCl, you need 1 mole of NaHCO3.
To calculate the amount of sodium bicarbonate needed, follow these steps:
Step 1: Convert the volume of HCl to moles.
To do this, we first need to calculate the number of moles of HCl using the given concentration and volume.
Moles of HCl = concentration (M) × volume (L)
= 6.00 mol/L × 0.225 L
= 1.35 moles of HCl
Step 2: Use the stoichiometry to find the moles of sodium bicarbonate.
From the balanced equation, we know that for every 1 mole of HCl, we need 1 mole of NaHCO3.
Moles of NaHCO3 = 1.35 moles of HCl
Step 3: Calculate the mass of sodium bicarbonate.
To do this, we need to know the molar mass of NaHCO3, which is 84.01 g/mol.
Mass of NaHCO3 = Moles of NaHCO3 × Molar mass of NaHCO3
= 1.35 moles × 84.01 g/mol
= 113.54 grams
Therefore, the mass of sodium bicarbonate that reacts with 225 mL of 6.00 M HCl is approximately 113.54 grams.
To find the mass of sodium bicarbonate (NaHCO3) that reacts with 225 mL of 6.00 M HCl, you need to follow these steps:
Step 1: Write a balanced chemical equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl):
NaHCO3 + HCl → NaCl + H2O + CO2
Step 2: Determine the moles of hydrochloric acid (HCl) using the given volume and concentration. Convert mL to L by dividing by 1000:
Volume of HCl = 225 mL = 225/1000 L = 0.225 L
Moles of HCl = volume (L) × concentration (M) = 0.225 L × 6.00 M = 1.35 moles
Step 3: Use the balanced chemical equation to determine the mole ratio between hydrochloric acid (HCl) and sodium bicarbonate (NaHCO3). From the equation, you can see that the ratio is 1:1.
Therefore, the moles of sodium bicarbonate (NaHCO3) required will also be 1.35 moles.
Step 4: Convert moles of sodium bicarbonate (NaHCO3) to the mass using its molar mass. The molar mass of sodium bicarbonate is the sum of the atomic masses of its elements:
Na: 22.99 g/mol
H: 1.01 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (x 3 because there are 3 oxygen atoms in NaHCO3)
Molar mass of NaHCO3 = 22.99 g/mol + 1.01 g/mol + 12.01 g/mol + (16.00 g/mol x 3) = 84.01 g/mol
Mass of sodium bicarbonate (NaHCO3) = moles × molar mass
Mass of NaHCO3 = 1.35 moles × 84.01 g/mol = 113.54 grams
Therefore, approximately 113.54 grams of sodium bicarbonate will react with 225 mL of 6.00 M HCl in the laboratory.