If 29.4 grams of hydrogen and 3701.3 grams of iodine combine to form hydrogen iodide, how many grams of hydrogen iodide must form?

Answer 1

Noah Aldrich

#3.730kg#

First write down a balanced chemical equation for the reaction :

#H_2+I_2->2HI#

This represents the mole ratio in which chemicals combine.

Now calculate the number of moles of each reactant present :

#n_(H_2)=m/M_r=29.4/2=14.7mol#

#n_(I_2)=m/M_r=3701.3/253.8=14.583mol#

Since the balanced chemical equation illustrates that the reactants combine in a #1:1# ratio, it implies that iodine is the limiting reagent and decides how much product is formed. Hydrogen is in excess.

So hence, since #1mol I_2# produces #2molHI#, it follows that #14.583molI_2# will produce #29.166molHI#

This corresponds to a mass of product of

#m=nxxM_r#

#=29.166xx127.9#

#=3730.33g#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer 2

Luna Chen

3730.7 grams of hydrogen iodide must form.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.