Starting with 195 g Li2O and 106 g H2O, decide which reactant is present in limiting quantities. Given: Li2O+H2O→2LiOH

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Taking into account the stoicionetry reaction and the definition of limiting reagent, Li₂O is present in limiting quantities.

The balanced reaction is:

Li₂O + H₂O → 2 LiOH

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Li₂O: 1 mole
H₂O: 1 mole
LiOH: 2 moles

The molar mass of each compound is:

Li₂O: 29.88 g/mole
H₂O: 18 g/mole
LiOH: 23.95 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Li₂O: 1 mole× 29.88 g/mole= 29.88 g
H₂O: 1 mole× 18 g/mole= 18 g
LiOH: 2 moles× 23.95 g/mole= 47.9 g

On the other side, the limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 18 grams of H₂O reacts with 29.88 grams of Li₂O, 106 grams of H₂O react with how much mass of Li₂O?

$mass of Li_{2} O=\frac{29.88 grams of Li_{2} Ox106 grams of H_{2} O}{18 grams of H_{2} O}$

mass of Li₂O= 175.96 grams

But 175.96 grams of Li₂O are not available, 195 grams are available. Since you have less moles than you need to react with 106 grams of H₂O, Li₂O will be the limiting reagent.

Finally, Li₂O is present in limiting quantities.

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Chlorine has two natural isotopes, chlorine-35 and chlorine-37.

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Answer:

The relative atomic mass of chlorine depends on the ratio between the abundance of these two naturally-occurring isotopes.

Explanation:

The relative atomic mass of an element is a weighted average of the atomic mass of its naturally-occurring isotopes. The relative abundance of each isotope gives its weight in this weighted average.

For these two naturally-occurring isotopes of chlorine:

The relative atomic mass of $^{35}{\rm Cl}$ is approximately $34.969$ daltons. The relative abundance of this isotope in nature is approximately $0.758$ .
The relative atomic mass $^{37}{\rm Cl}$ is approximately $36.966$ daltons. The relative abundance of this isotope in nature is approximately $0.242$ .

$\begin{array}{|c|c|c|}\cline{1-3} \text{Isotope} & \text{Atomic Mass} & \text{Relative Abundance}\\ \cline{1-3} ^{35}{\rm Cl} & \approx 34.968\; \rm Da} & \approx 0.758 \\ \cline{1-3} ^{37}{\rm Cl} & \approx 36.966\; \rm Da & \approx 0.242 \\ \cline{1-3}\end{array}$ .

$\begin{aligned}&\text{relative atomic mass of Cl} \\ &= \text{atomic mass of $^{35}{\rm Cl}$} \times \text{relative abundance of $^{35}{\rm Cl}$} \\&\quad + \text{atomic mass of $^{37}{\rm Cl}$} \times \text{relative abundance of $^{37}{\rm Cl}$} \\ &\approx 34.968\; \rm Da \times 0.758 + 36.966\; \rm Da \times 0.242 \\ &\approx 35.45\; \rm Da \end{aligned}$ .

The relative abundance of $^{35}{\rm Cl}$ is much higher than that of $^{37}{\rm Cl}$ . Consequently, the relative atomic mass of the element $\rm Cl$ is closer to the atomic mass of $^{35}{\rm Cl}\!$ than that of $^{37}{\rm Cl}\!$ .