# What is quicklime? And what occurs when a 10*g mass of this material reacts with the carbon dioxide liberated from the decomposition of 1.80*g of glucose?

Aug 3, 2017

Well, $\text{quick lime}$ is simply $C a O \left(s\right)$........and we get a $10 \cdot g$ mass of $\text{calcium carbonate.........}$

#### Explanation:

And we look at the 1:1 reaction.......

$C a O \left(s\right) + C {O}_{2} \left(g\right) \stackrel{\text{water}}{\rightarrow} C a C {O}_{3} \left(s\right)$ $\left(i\right)$

$\text{Moles of calcium oxide} \equiv \frac{5.6 \cdot g}{56.08 \cdot g \cdot m o {l}^{-} 1} = 0.10 \cdot m o l$

And glucose combusts according to ..........

${C}_{6} {H}_{12} {O}_{6} \left(s\right) + 6 {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right)$ $\left(i i\right)$

And thus moles of carbon dioxide liberated......

$\frac{1.80 \cdot g}{180.16 \cdot g \cdot m o {l}^{-} 1} \times 6 \equiv 0.600 \cdot m o l$

And thus because we have EXCESS carbon dioxide, $C {O}_{2}$ is the reagent in excess in $\left(i\right)$...........

And thus we gets $0.10 \cdot m o l$ $C a C {O}_{3}$, which represents a mass of $0.10 \cdot m o l \times 100.09 \cdot g \cdot m o {l}^{-} 1 = 10.0 \cdot g$ $\text{calcium carbonate}$