Example: serum containing na gave a signal of 4.27 mv in an atomic emission analysis. then, 5 ml of 2.08 m nacl were added to 95.0 ml of serum. the spiked serum gave a signal of 7.98 mv. find the original concentration of nat in serum.

Explanation:

It is known that,

       No. of moles = Concentration × volume

As 5 ml of 2.08 M NaCl NaS added to 95 ml of serum then,

        Total volume = (95 + 5) ml = 100 ml

Therefore, total moles of = (5 × 2.08) + (95 × x)

where,      x =  concentration of in original serum

Hence, signal obtained for this NaS = 7.8 mV

Let us assume same volume, that is, 100 ml Serum then will be (100 × x) mol.

Signal obtained for this is 4.27 mV.

Now, the ratio of signal is equal to the ratio of mole.

So,  

                           x = 0.113

Thus, we can conclude that original concentration of in Serum is 0.113.

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

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