. How many grams of H2SO4 (98.1 g·mol-1) are required to react with 4.0 g NaOH (40.0 g·mol-1) in this neutralization reaction?H2SO4 + 2NaOH Na2SO4 + 2H2O a. 2.4 g
b. 4.9 g
c. 9.8 g
d. 20 g
Here is a worked example.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To find the number of grams of H2SO4 required to react with 4.0 g of NaOH, we need to use stoichiometry and the molar ratios from the balanced equation:
H2SO4 + 2NaOH -> Na2SO4 + 2H2O
First, calculate the number of moles of NaOH using the formula:
moles = mass / molar mass
moles of NaOH = 4.0 g / 40.0 g·mol-1 = 0.1 mol
From the balanced equation, we can see that the molar ratio between NaOH and H2SO4 is 2:1. This means that for every 2 moles of NaOH, we need 1 mole of H2SO4.
So, to determine the number of moles of H2SO4 needed, divide the number of moles of NaOH by 2:
moles of H2SO4 = 0.1 mol / 2 = 0.05 mol
Now, we can calculate the mass of H2SO4 using the formula:
mass = moles * molar mass
mass of H2SO4 = 0.05 mol * 98.1 g·mol-1 = 4.91 g
Therefore, the answer is approximately 4.91 g. Therefore, the correct answer is b. 4.9 g.