Solutions of sulfuric acid and lead(II) acetate react to form solid lead(II) sulfate and a solution of acetic acid. 4.90 g of sulfuric acid and 4.90 g of lead(II) acetate are mixed. Calculate the number of grams of sulfuric acid, lead(II) acetate, lead(II) sulfate, and acetic acid present in the mixture after the reaction is complete .

Mass H2SO4 = 3.42 grams

Mass of lead acetate = 0 grams

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 1.81 grams

Explanation:

Step 1: Data given

Mass of sulfuric acid = 4.90 grams

Molar mass of sulfuric acid = 98.08 g/mol

Mass of lead acetate = 4.90 grams

Molar mass of lead acetate = 325.29 g/mol

Step 2: The balanced equation

H2SO4 + Pb(C2H3O2)2 → PbSO4 + 2CH3COOH

Step 3: Calculate moles

Moles = mass / molar mass

Moles H2SO4 = 4.90 grams / 98.08 g/mol

Moles H2SO4 = 0.0500 moles

Moles lead acetate = 4.9 grams / 325.29 g/mol

Moles lead acetate = 0.0151 moles

Step 4: Calculate the limiting reactant

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

The limiting reactant is lead acetate. It will completzly be consumed (0.0151 moles). H2SO4 is in excess. There will react 0.0151 moles. There will remain 0.0500 - 0.0151 = 0.0349 moles

Step 5: Calculate moles of products

For 1 mol H2SO4 we need 1 mol lead acetate to produce 1 mol PbSO4 and 2 moles CH3COOH

For 0.0151 moles lead acetate we'll have 0.0151 moles PbSO4 and 2*0.0151 = 0.0302 moles CH3COOH

Step 6: Calculate mass

Mass = moles * molar mass

Mass H2SO4 = 0.0349 moles * 98.08 g/mol

Mass H2SO4 = 3.42 grams

Mass PbSO4 = 0.0151 moles * 303.26 g/mol

Mass PbSO4 = 4.58 grams

Mass of CH3COOH = 0.0302 moles * 60.05 g/mol

Mass of CH3COOH = 1.81 grams

a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.