2 LiOH(s) + CO2(g) → Li2CO3(s) + H2O(l) An astronaut exhales 880 g of CO2(g) per day. According to t — Stoichiometry Chemistry Question

STEPS:

1. Identify the given values, molar masses, and chemical equation:
* Reaction: $\text{CO}_2(g) + 2\ \text{LiOH}(s) \rightarrow \text{Li}_2\text{CO}_3(s) + \text{H}_2\text{O}(l)$
* Mass of $\text{CO}_2$ exhaled: $880\text{ g}$
* Molar mass of $\text{CO}_2$: $44\text{ g/mol}$
* Molar mass of $\text{LiOH}$: $24\text{ g/mol}$

2. Convert the mass of $\text{CO}_2$ into moles:
To determine how many moles of carbon dioxide the astronaut produces in one day, divide the given mass by its molar mass:
$$\text{Moles of } \text{CO}_2 = \frac{880\text{ g}}{44\text{ g/mol}} = \mathbf{20\text{ mol of } \text{CO}_2} \quad$$

3. Relate reactant moles using the stoichiometric ratio:
Examine the coefficients in the balanced chemical equation. The stoichiometric ratio is $2\text{ moles of LiOH}$ for every $1\text{ mole of CO}_2$. Use this ratio as a conversion factor to find the moles of $\text{LiOH}$ required to completely consume the carbon dioxide:
$$20\text{ mol of } \text{CO}_2 \times \frac{2\text{ mol of LiOH}}{1\text{ mol of } \text{CO}_2} = \mathbf{40\text{ mol of LiOH}} \quad$$

4. Convert the required moles of $\text{LiOH}$ back into grams:
Multiply the calculated moles of lithium hydroxide by its molar mass to find the final mass required:
$$40\text{ mol of LiOH} \times 24\text{ g/mol} = \mathbf{960\text{ g of LiOH}} \quad$$
This calculation successfully identifies Option D as the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect (240 g): This value is obtained if a student makes a dual mathematical error—first failing to apply the $2:1$ stoichiometric ratio of reactants, and then mistakenly dividing by 2 instead of multiplying by 2 (calculating $10\text{ moles} \times 24\text{ g/mol} = 240\text{ g}$).
• Option B is incorrect (480 g): A student will arrive at this number if they completely overlook the $2:1$ stoichiometric coefficient ratio of $\text{LiOH}$ to $\text{CO}_2$ in the balanced equation. Calculating a $1:1$ mole ratio yields $20\text{ moles of LiOH}$, which translates to $20\text{ mol} \times 24\text{ g/mol} = 480\text{ g}$.
• Option C is incorrect (720 g): This value corresponds to $30\text{ moles of LiOH}$ ($30\text{ mol} \times 24\text{ g/mol} = 720\text{ g}$), which would only result if the student incorrectly applied a $1.5:1$ mole ratio instead of the correct $2:1$ ratio.