what mass of KOH is reqired to react completely with 2.70g of H2So4 to produce K2S04 and water?
To find the mass of KOH required to react completely with 2.70g of H2SO4, we need to use the balanced chemical equation for the reaction between KOH and H2SO4.
The balanced chemical equation is:
2 KOH + H2SO4 -> K2SO4 + 2 H2O
From the equation, we can see that 2 moles of KOH react with 1 mole of H2SO4. We need to convert the given mass of H2SO4 (2.70g) to moles using its molar mass before determining the required mass of KOH.
1. Calculate the molar mass of H2SO4:
H2 - 2(1.0079 g/mol) = 2.0158 g/mol
S - 1(32.06 g/mol) = 32.06 g/mol
O4 - 4(16.00 g/mol) = 64.00 g/mol
Molar mass of H2SO4 = 2(2.0158) + 32.06 + 4(16.00) = 98.09 g/mol
2. Calculate the number of moles of H2SO4:
moles = mass / molar mass
moles = 2.70g / 98.09 g/mol ≈ 0.0275 mol
3. Convert the moles of H2SO4 to moles of KOH (using the mole ratio from the balanced equation):
moles of KOH = 2 * moles of H2SO4 ≈ 2 * 0.0275 mol = 0.055 mol
4. Calculate the mass of KOH:
mass = moles * molar mass
mass = 0.055 mol * 56.11 g/mol ≈ 3.08 g
Therefore, approximately 3.08 grams of KOH is required to react completely with 2.70 grams of H2SO4 to produce K2SO4 and water.
Just follow the steps on the link below.
http://www.jiskha.com/science/chemistry/stoichiometry.html