Chemistry: If you start with 4.00 moles of c3h8 (propane) and 4.00 moles of o2, how many moles of carbon dioxide can be produced?

If you start with 4.00 moles of c3h8 (propane) and

breannaasmith1122

09.06.2020

Answer : The correct option is, 5

Solution : Given,

Mass of nickel-59 = 0.17 g

Mass of cobalt-59 = 5.27 g

Equation for the radioactive decay of nickel-59 is :

$_{28}^{59}\textrm{Ni}\rightarrow _{27}^{59}\textrm{Co}+_{-1}^{0}\beta$

Now, we have to calculate the initial amount of nickel-59, we are using the stoichiometry of the reaction and moles of the reactant and product.

Formula used : $\text{Moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of $_{28}^{59}\textrm{Ni}=\frac{0.17g}{59g/mol}=0.00288moles$

Moles of $_{27}^{59}\textrm{Co}=\frac{5.27g}{59g/mol}=0.089moles$

By stoichiometry of the reaction,

1 mole of $_{27}^{59}\textrm{Co}$ is produced by 1 mole $_{28}^{59}\textrm{Ni}$

So, 0.089 moles of $_{27}^{59}\textrm{Co}$ will be produced by = $\frac{1}{1}\times 0.089=0.089\text{ moles of }_{28}^{59}\textrm{Ni}$

Amount of $_{28}^{59}\textrm{Ni}$ decomposed will be = 0.089 moles

Initial amount of $_{28}^{59}\textrm{Ni}$ will be = Amount decomposed + Amount left = (0.00288 + 0.089)moles = 0.09188 moles

Now, to calculate the number of half lives, we use the formula :

$a=\frac{a_o}{2^n}$

where,

a = amount of reactant left after n-half lives = 0.00288 moles

a_o = Initial amount of the reactant = 0.09188 moles

n = number of half lives

Now put all the given values in above equation, we get

$0.00288=\frac{0.09188}{2^n}$

2^n=32.81

Taking log on both sides, we get

$n\log2=\log(32.81)\\n=5.03=5$

Therefore, '5' number of half-lives have passed since the meteorite formed .

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