Question

Given the following reaction:

H2SO4 + 2LiOH --> Li2SO4 + 2H2O


What mass of water is produced from 19 g of sulfuric acid?

Question options:

1.0


7.0


2.0


6.3

Answer 1

The mass of water produced from 19 g of sulfuric acid reacting with lithium hydroxide is approximately 7.0 g when calculated using mole-mass stoichiometry.

Step-by-step explanation:

The question asks what mass of water is produced from 19 g of sulfuric acid (H2SO4) reacting with lithium hydroxide (LiOH) to produce lithium sulfate (Li2SO4) and water (H2O). To find the mass of water produced, we use mole-mass stoichiometry.

First, calculate the number of moles of H2SO4 using its molar mass:

• Molar mass of H2SO4 = 2(1.01 g/mol) + 32.07 g/mol + 4(16.00 g/mol) = 98.09 g/mol
• Moles of H2SO4 = 19 g / 98.09 g/mol ≈ 0.1937 mol

According to the balanced equation, one mole of H2SO4 produces two moles of H2O. Hence, we have:

• Moles of H2O = 0.1937 mol H2SO4 × 2 mol H2O/mol H2SO4 ≈ 0.3874 mol H2O

Now, calculate the mass of water produced:

• Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
• Mass of H2O = 0.3874 mol × 18.02 g/mol ≈ 6.98 g, which rounds to 7.0 g.

Answer 2

Approx. 4⋅g.

Step-by-step explanation:

Moles of sulfuric acid =10.0⋅g98.08⋅g⋅mol−1=0.102⋅mol.

Now we have the molar quantity of sulfuric acid that react; we also have the stoichiometric equation that shows the molar equivalence of sulfuric acid, and lithium hydroxide.

Given the stoichiometry,

mass of water =0.102⋅mol×2×18.01.g.mol−1=??⋅g.

Why did I multiply the mass in this equation by 2? Am I pulling your leg?