how many grams of the excess reactant are left over according to the reaction below given that you start with 10.0 g of al and 19.0 grams of o2?

divide 10.0g by each reactant's respective molar mass then divide that number by its coefficient in the balanced reaction. the smaller of the two numbers indicates the limiting reactant - it’s Al

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.