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3 years ago

C12H22O11 + 12O2 ---> 12CO2 + 11H2O

1 answer:

5 0

Molar mass sucrose ( C12H22O11 ) = 342 g/mol

8.0 g of O2 is the limiting factor of the reaction reagent

C12H22O11 + 12 O2 ------------> 12 CO2 + 11 H2O

1 mol C12H22O11 ------------- 342 g
x mol C12H22O11 ------------- 10 g

x = 10 / 342

x = 0,029 mol of sucrose

answer A

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3 years ago

If 45.00 g of precipitate is formed from the reaction of 0.100 mol/L

Tom [10]

Answer:

Approximately $2.53\; \rm L$ (rounded to three significant figures) assuming that ${\rm HCl}\, (aq)$ is in excess.

Explanation:

When ${\rm HCl} \, (aq)$ and ${\rm AgNO_3}\, (aq)$ precipitate, ${\rm AgCl} \, (s)$ (the said precipitate) and $\rm HNO_3\, (aq)$ are produced:

${\rm HCl}\, (aq) + {\rm AgNO_3}\, (aq) \to {\rm AgCl}\, (s) + {\rm HNO_3}\, (aq)$ (verify that this equation is indeed balanced.)

Look up the relative atomic mass of $\rm Ag$ and $\rm Cl$ on a modern periodic table:

$\rm Ag$ : $107.868$ .
$\rm Cl$ : $35.45$ .

Calculate the formula mass of the precipitate, $\rm AgCl$ :

$\begin{aligned}& M({\rm AgCl})\\ &= (107.868 + 35.45)\; \rm g \cdot mol^{-1} \\\ &\approx 143.318 \; \rm g\cdot mol^{-1}\end{aligned}$ .

Calculate the number of moles of $\rm AgCl$ formula units in $45.00\; \rm g$ of this compound:

$\begin{aligned}n({\rm AgCl}) &= \frac{m({\rm AgCl})}{M({\rm AgCl})} \\ &\approx \frac{45.00\; \rm g}{143.318\; \rm g \cdot mol^{-1}}\approx 0.313987\; \rm mol \end{aligned}$ .

Notice that in the balanced equation for this reaction, the coefficients of ${\rm AgNO_3} \, (aq)$ and ${\rm AgCl}\, (s)$ are both one.

In other words, if ${\rm HCl}\, (aq)$ (the other reactant) is in excess, it would take exactly $1\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $1\; \rm mol \!$ of ${\rm AgCl}\, (s)\!$ formula units.

Hence, it would take $0.313987\; \rm mol$ of ${\rm AgNO_3} \, (aq)\!$ formula units to produce $0.313987\; \rm mol\!$ of ${\rm AgCl}\, (s)\!$ formula units.

Calculate the volume of the ${\rm AgNO_3} \, (aq)\!$ solution given that the concentration of the solution is $0.124\; \rm mol \cdot L^{-1}$ :

$\begin{aligned}V({\rm AgNO_3}) &= \frac{n({\rm AgNO_3})}{c({\rm AgNO_3})} \\ &\approx \frac{0.313987\; \rm mol}{0.124\; \rm mol \cdot L^{-1}}\approx 2.53\; \rm L\end{aligned}$ .

(The answer was rounded to three significant figures so as to match the number of significant figures in the concentration of ${\rm AgNO_3} \, (aq)\!$ .)

In other words, approximately $2.53\; \rm L$ of that ${\rm AgNO_3} \, (aq)\!$ solution would be required.

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3 years ago

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