How many milliliter of 0.150 M H2SO4 (sulfuric acid) are required to react with 1.87 g of sodium hydrogen carbonate, NaHCO3, according to the following equation?
H2SO4 + 2NaHCO3 --> Na2SO4 + 2H2O + 2CO2
dont understand this am i suppose to change 1.87 into Molarity for liters or something?? never learned how to do that
thanks for the comments in advance
Grams don't convert to liters or to molarity.. Grams converts to moles. moles = grams/molar mass. Then you use the equation to convert moles of one material to moles of the other. If you have M and L, then moles = M x L.
So just follow the steps.
Convert 1.87 grams NaHCO3 to moles. moles = grams/molar mass.
2. You have moles NaHCO3, now convert that to moles H2SO4 using the coefficients in the balanced equation.
3. Now you have moles H2SO4. M x L = moles. You have moles and M from the problem, solve for L and convert to mL.
@DrBob222 the link you provided doesnt do a good job in explaining how to convert grams into liters/Molarity
if you can do u think u can go into a little more detail on showing me what to do with the 1.87 sodium hydrogen carbonate?
thanks
Well, chemistry can be quite confusing sometimes, but don't worry, I'm here to help, even if it means cracking a few jokes along the way!
To solve this problem, you'll need to convert the mass of sodium hydrogen carbonate (NaHCO3) into moles. Then, you can use the molar ratio from the balanced chemical equation to determine the number of moles of sulfuric acid (H2SO4) needed. And finally, you can convert moles of H2SO4 into milliliters.
Now, let's dive in! But be warned, we might get a little "salty" along the way!
Step 1: Convert the mass of NaHCO3 to moles.
Molar mass of NaHCO3:
Na = 22.99 g/mol
H = 1.01 g/mol
C = 12.01 g/mol
O = 16.00 g/mol (3 oxygen atoms)
Total molar mass = 22.99 + 1.01 + 12.01 + (16.00 x 3) = 84.00 g/mol
So, moles of NaHCO3 = mass (in grams) / molar mass (in grams per mole)
Step 2: Use the molar ratio to find the moles of H2SO4 needed.
From the balanced equation, we see that 2 moles of NaHCO3 react with 1 mole of H2SO4.
Step 3: Convert moles of H2SO4 to milliliters.
To do this, you need to know the molarity (0.150 M) of the sulfuric acid solution.
We're almost there! Just remember that 1 liter = 1000 milliliters.
Use the equation Molarity (M) = moles / volume (in liters) to calculate the volume (in liters) of H2SO4. Then, multiply it by 1000 to convert to milliliters.
And voila! You've got your answer! But hey, don't get too "salty" about all these calculations. Remember, laughter is always the best solution, even in chemistry!
Hope this helps! Let me know if you need any more assistance, or if you just want another joke to lighten the mood!
To determine the volume of 0.150 M H2SO4 required to react with 1.87 g of NaHCO3, we need to follow a few steps.
First, let's calculate the number of moles of NaHCO3 using its molar mass. The molar mass of NaHCO3 is the sum of the atomic masses of sodium (Na), hydrogen (H), carbon (C), and three oxygen (O) atoms, which is approximately 84.01 g/mol.
Number of moles of NaHCO3 = (mass of NaHCO3) / (molar mass of NaHCO3)
= 1.87 g / 84.01 g/mol
Next, let's use stoichiometry to determine the moles of H2SO4 required to react with the calculated moles of NaHCO3. From the balanced chemical equation, we know that the stoichiometric ratio between H2SO4 and NaHCO3 is 1:2.
Number of moles of H2SO4 = 2 * (number of moles of NaHCO3)
Now that we have the number of moles of H2SO4, we can use the molarity (M) to find the volume of H2SO4 needed. Molarity is defined as moles of solute per liter of solution (mol/L).
Volume of H2SO4 = (number of moles of H2SO4) / (molarity of H2SO4)
= (2 * number of moles of NaHCO3) / (0.150 mol/L)
Finally, since the molarity is in moles per liter (mol/L), we need to convert the volume of H2SO4 from liters to milliliters by multiplying by 1000.
Volume of H2SO4 (in milliliters) = Volume of H2SO4 (in liters) * 1000
By following these steps, you should be able to calculate the volume of 0.150 M H2SO4 needed to react with 1.87 g of NaHCO3.
A step by step procedure for working these problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html