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Sagot :
Answer: a. [tex]2CH_4+2NH_3+3O_2\rightarrow 2HCN+6H_2O(g)[/tex]
b. [tex]CH_4[/tex] is the limiting.
c. 13.5 g of hydrogen cyanide will be formed
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of methane}=\frac{8g}{16g/mol}=0.50moles[/tex]
[tex]\text{Moles of ammonia}=\frac{10g}{17g/mol}=0.59moles[/tex]
The balanced reaction will be :
[tex]2CH_4+2NH_3+3O_2\rightarrow 2HCN+6H_2O(g)[/tex]
According to stoichiometry :
2 moles of [tex]CH_4[/tex] require = 2 moles of [tex]NH_3[/tex]
Thus 0.5 moles of [tex]CH_4[/tex] will require=[tex]\frac{2}{2}\times 0.5=0.5moles[/tex] of [tex]NH_3[/tex]
Thus [tex]CH_4[/tex] is the limiting reagent as it limits the formation of product and [tex]NH_3[/tex] is the excess reagent as it is present in more amount than required.
As 2 moles of [tex]CH_4[/tex] give = 2 moles of [tex]HCN[/tex]
Thus 0.5 moles of [tex]CH_4[/tex] give =[tex]\frac{2}{2}\times 0.5=0.5moles[/tex] of [tex]HCN[/tex]
Mass of [tex]HCN=moles\times {\text {Molar mass}}=0.5moles\times 27g/mol=13.5g[/tex]
Thus 13.5 g of [tex]HCN[/tex] will be produced from the given masses of both reactants.
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