Respuesta :
Answer :
524.7 grams
Explanation:
this question depend on the limiting reactant calculation where we should determine the mass that consumed completely during the reaction
we determine it by calculate the number of moles for each compound and compared to the moles of the balanced equation
n HI = mass / molar mass
= 481/ 127 = 3.79 moles
n KHCO3 = 318/100 = 3.18 moles ( the rate of the moles in the balanced equation is 1:1
the limiting reactant product is the least amount KHCO3
KHCO3 HI
100g/mole 165 g/mole
318 g × solving it u find x = 524.7 g of KI
Answer:
Approximately [tex]527\; \rm g[/tex].
Explanation:
Look up relative atomic mass data on a modern periodic table:
- [tex]\rm H[/tex]: [tex]1.008[/tex].
- [tex]\rm I[/tex]: [tex]126.904[/tex].
- [tex]\rm K[/tex]: [tex]39.098[/tex].
- [tex]\rm C[/tex]: [tex]12.011[/tex].
- [tex]\rm O[/tex]: [tex]15.999[/tex].
Calculate the formula mass of [tex]\rm HI[/tex] and [tex]\rm KHCO_3[/tex] (the two reactants,) as well as [tex]\rm KI[/tex] (the product.)
- [tex]M(\mathrm{HI}) \approx 1.008 + 126.904 =127.912\; \rm g \cdot mol^{-1}[/tex].
- [tex]M(\mathrm{KHCO_3}) \approx 39.098 + 1.008 + 12.011 + 3 \times 15.999 = 100.114\; \rm g \cdot mol^{-1}[/tex].
- [tex]M(\mathrm{KI}) \approx 39.098 + 126.904 = 166.002\; \rm g \cdot mol^{-1}[/tex]
The question states that [tex]m(\mathrm{HI}) = 481\; \rm g[/tex] while [tex]m(\mathrm{KHCO_3}) = 318\; \rm g[/tex]. Calculate the number of moles of formula units in each of these two reactants:
[tex]\begin{aligned} n(\mathrm{HI}) &= \frac{m(\mathrm{HI})}{M(\mathrm{HI})} \\ &\approx \frac{481\; \rm g}{127.912\; \rm g \cdot mol^{-1}} \approx 3.76\; \rm mol\end{aligned}[/tex].
[tex]\begin{aligned} n(\mathrm{KHCO_3}) &= \frac{m(\mathrm{KHCO_3})}{M(\mathrm{KHCO_3})} \\ &\approx \frac{318\; \rm g}{100.114\; \rm g \cdot mol^{-1}} \approx 3.18\; \rm mol\end{aligned}[/tex].
Note that in the balanced equation of this reaction, the ratio between the coefficients of [tex]\rm HI[/tex], [tex]\rm KHCO_3[/tex], and [tex]\rm KI[/tex] is [tex]1:1:1[/tex].
Therefore, for every mole of [tex]\rm HI[/tex] formula units consumed, one mole of [tex]\rm KHCO_3[/tex] formula units will also be consumed, while one mole of [tex]\rm KI[/tex] formula units will be produced.
Similarly, for every mole of [tex]\rm KHCO_3[/tex] formula units consumed, one mole of [tex]\rm HI[/tex] formula units will also be consumed, while one mole of [tex]\rm KI[/tex] formula units will be produced.
Assume that [tex]\rm HI[/tex] is in excess. If all these (approximately) [tex]3.76\; \rm mol[/tex] of [tex]\rm HI\![/tex] formula units are consumed, [tex]3.76\; \rm mol\![/tex] of [tex]\rm KHCO_3[/tex] formula units will also need to be consumed. However, that's not possible because there was only approximately [tex]3.18\; \rm mol[/tex] of [tex]\rm KHCO_3\![/tex] formula units available.
On the other hand, if [tex]\rm KHCO_3[/tex] is in excess, all these (approximately) [tex]3.18\; \rm mol[/tex] moles of [tex]\rm KHCO_3[/tex] formula units would be consumed. At the same time, approximately [tex]3.18\; \rm mol\![/tex] of [tex]\rm HI[/tex] would be consumed- which is indeed possible because approximately [tex]3.76\; \rm mol[/tex] of [tex]\rm HI\![/tex] formula units are available. Therefore, up to approximately [tex]3.18\; \rm mol\!\![/tex] of [tex]\rm KI[/tex] formula units will be produced.
Calculate the mass of that [tex]3.18\; \rm mol\!\![/tex] of [tex]\rm KI[/tex] formula units:
[tex]\begin{aligned}m(\mathrm{KI}) &= n(\mathrm{KI}) \cdot M(\mathrm{KI}) \\ &\approx 3.18\; \rm mol \times 166.002\; \rm g \cdot mol^{-1} \approx 527\; \rm g\end{aligned}[/tex].
