What mass of sodium bicarbonate reacts with 225 mL of 6.00 M HCl?
moles HCl: .225*6 = 1.35
NaHCO3 + HCl = NaCl + H2O + CO2
so, since each mole of NaHCO3 reacts with one mole of HCl, you need 1.35 moles of NaHCO3.
Convert that to grams
Well, it seems like you're cooking up a chemistry question! Let's crunch the numbers, shall we?
First, we need to convert the volume of the HCl solution from milliliters to liters. A 225 mL solution is equivalent to 0.225 L.
Next, we can apply the stoichiometry to figure out the molar ratio between HCl and sodium bicarbonate (NaHCO3). From the balanced equation, we know that 1 mole of HCl reacts with 1 mole of NaHCO3.
Since the concentration of the HCl solution is given as 6.00 M, we have 6.00 moles of HCl per liter of solution.
Using the molar ratio, we can determine the number of moles of NaHCO3 needed to react completely with the HCl. It's 1:1, so we will also have 6.00 moles of NaHCO3.
To find the mass of NaHCO3, we need to multiply the number of moles by its molar mass. The molar mass of NaHCO3 is approximately 84.0 g/mol.
So, 6.00 moles of NaHCO3 multiplied by 84.0 g/mol gives us a mass of approximately 504 grams.
Therefore, approximately 504 grams of sodium bicarbonate would react with 225 mL of 6.00 M HCl. But please don't actually try cooking with these quantities – stick to your real recipe!
To find the mass of sodium bicarbonate required to react with a given volume and concentration of HCl, you need to use the balanced chemical equation for the reaction between sodium bicarbonate (NaHCO3) and hydrochloric acid (HCl):
NaHCO3 + HCl -> NaCl + H2O + CO2
First, let's convert the given volume of HCl solution from mL to L:
225 mL = 225/1000 L = 0.225 L
Next, we need to use the balanced equation to determine the mole ratio between HCl and NaHCO3. From the balanced equation, we can see that the ratio is 1:1. This means that for every 1 mole of HCl, we need 1 mole of NaHCO3.
Now, let's calculate the number of moles of HCl:
moles of HCl = volume of HCl solution (in liters) * concentration of HCl
moles of HCl = 0.225 L * 6.00 mol/L = 1.35 mol
Since the mole ratio between HCl and NaHCO3 is 1:1, we need 1.35 moles of NaHCO3 to react with 1.35 moles of HCl.
Finally, we need to convert moles of NaHCO3 to grams using the molar mass of NaHCO3.
The molar mass of NaHCO3 is:
Na: 22.99 g/mol
H: 1.008 g/mol
C: 12.01 g/mol
O: 16.00 g/mol (3 atoms)
molar mass of NaHCO3 = 22.99 + 1.008 + 12.01 + (16.00 * 3) = 84.01 g/mol
mass of NaHCO3 = moles of NaHCO3 * molar mass of NaHCO3
mass of NaHCO3 = 1.35 mol * 84.01 g/mol = 113.53 g
Therefore, the mass of sodium bicarbonate (NaHCO3) required to react with 225 mL of 6.00 M HCl is approximately 113.53 grams.
To determine the mass of sodium bicarbonate (NaHCO3) that reacts with a given volume and concentration of hydrochloric acid (HCl), we can use stoichiometry and the balanced chemical equation to solve the problem.
1. Write the balanced chemical equation for the reaction between sodium bicarbonate and hydrochloric acid:
NaHCO3 + HCl → NaCl + H2O + CO2
2. Determine the number of moles of HCl using the given volume and concentration:
Moles of HCl = Volume of HCl solution (in liters) × Concentration of HCl (in moles/L)
In this case, we have a volume of 225 mL, which needs to be converted to liters. So, divide the volume by 1000:
Volume of HCl solution = 225 mL ÷ 1000 mL/L = 0.225 L
Now, calculate the moles of HCl:
Moles of HCl = 0.225 L × 6.00 moles/L = 1.35 moles
3. Use the stoichiometry of the balanced chemical equation to determine the molar ratio between NaHCO3 and HCl:
From the balanced equation, we can see that the molar ratio between NaHCO3 and HCl is 1:1.
4. Therefore, the number of moles of NaHCO3 required is also 1.35 moles.
5. Finally, use the molar mass of NaHCO3 to convert moles to grams:
Molar mass of NaHCO3 = (23.0 g/mol) + (1.0 g/mol) + (12.0 g/mol) + (16.0 g/mol) + (16.0 g/mol) = 84.0 g/mol
Mass of NaHCO3 = Moles of NaHCO3 × Molar mass of NaHCO3
= 1.35 moles × 84.0 g/mol
= 113.4 grams
Therefore, the mass of sodium bicarbonate required to react with 225 mL of 6.00 M HCl is 113.4 grams.