An unknown compound is discovered in a raid on a terrorist organization; it is believed that the compound is sarin. when a 10.0-gram sample of this compoun is completely combusted, 15.6 g co2 and 6.4 g h20 are produced, along with other combustion products. using numerical calculations, prove that this unknown compound can not be sarin.

Given:

Mass of the unknown compound (believed to be Sarin) = 10.0 g

Mass of CO2 produced = 15.6 g

Mass of H2O produced = 6.4 g

To determine:

If the unknown compound is Sarin

Explanation:

The molecular formula for Sarin is: C₄H₁₀FO₂P

Based on stoichiometry the C:H ratio = 4:10 i.e 2:5

Now, from the combustion data:

# moles of CO2 produced = # moles of C present in the unknown sample

Similarly, (# moles of H2O produced)*2 = # moles of H present in the unknown sample

# moles of CO2 = 15.6 g/44 g.mol-1 = 0.354 moles

# moles of H2O = (6.4 g/ 18 g.mol-1 )* 2 = 0.711 moles

The C:H ratio = 0.354 : 0.711 = 1:2

Ans: Since the C:H ratio does not match the actual stoichiometry based on the molecular formula, the unknown compound is not Sarin.

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons

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