Answer to Question #344969 in General Chemistry for gle

Question

For the following reaction, 4.41 grams of hydrogen gas are allowed to react with 39.8 grams of iodine .

hydrogen(g) + iodine(s)  hydrogen iodide(g)

What is the maximum mass of hydrogen iodide that can be formed?  grams

What is the FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?  grams

Answer 1

H₂ + I₂ = 2HI

n(H₂) = m(H₂)/Mr(H₂) = 4.41 g / 2 g/mol = 2.2 moles

n(I₂) = m(I₂) / Mr(I₂) = 39.8 g / 254 g/mol = 0.16 moles

H₂ - excess reagent

I₂ - limiting reagent

m(HI) = (m(I₂) * 2Mr(HI)) / Mr(I₂) = (39.8 g * 2*128 g/mol) / 254 g/mol = 40.11 g

m(H₂) = (m(I₂)*Mr(H₂)) / Mr(I₂) = (39.8 g * 2 g/mol) / 254 g/mol = 0.31 g

after a the reaction m(H₂) = 4.41 g - 0.31 g = 4.1 g