Question #7a6ab

#"33.3 g HI"# will be produced.

Start with a balanced equation:

#"H"_2("g")# + #"I"_2(s)# #rarr# #"2HI(g)"#

Convert 33.0 grams of iodine to moles of iodine by dividing the given mass by its molar mass. The molar mass of #"I"_2# is 253.80894 g/mol.

#"33.0 g I"_2"##-:##"253.80894 g/mol"# = #"0.13001 mol I"_2#

From the balanced equation, the mole ratio of #"I"_2# to #"HI"# is #"1 mol I"_2": 2 mol I"_2#.

Determine the number of moles of #"HI"# that can be produced from #"0.13001 mol I"_2# by multiplying times the mole ratio so that #"HI"# is on top.

#"0.13001 mol I"_2# x #"2 mol HI"/"1 mol I2"# = #"0.26002 mol HI"#

Calculate the mass of #"HI"# in grams by multiplying the mol #"HI"# times its molar mass. The molar mass of #"HI"# is #"127.91247 g/mol"#.

#"0.26002 mol HI"# x #"127.91247 g HI"/"1 mol HI"# = #"33.259 g HI"# = #"33.3g HI"# due to 3 sig figs in 33.0 g of #"I"_2"# stated in the problem.

33.26 g of HI will be formed.

#H_(2(g))+I_(2(g))rarr2HI_((g))#

1mol + 1mol #rarr# 2mol

Using approximate #A_r# values:

#(2xx127)gI_2rarr2(1+127)gHI#

So:

#254grarr256g#

So:

#1grarr256/254g#

So:

#33.0grarr(256)/(254)xx33.0= 33.26g#