Question

environmental sensors detect 3.42 grams of h₂o inside the lunar module. how many moles of lioh are consumed to produce this much h₂o? show all of your work. you must be able to prove your answer. 0.190 moles 0.380 moles 1.00 moles 18.02 moles

Explanation:

Step1: Find molar mass of \(H_2O\)

Molar mass of \(H = 1.008\ g/mol\), \(O = 16.00\ g/mol\).
Molar mass of \(H_2O = 2\times1.008 + 16.00 = 18.016\ g/mol\approx18.02\ g/mol\).

Step2: Calculate moles of \(H_2O\)

Moles \(=\frac{\text{mass}}{\text{molar mass}}=\frac{3.42\ g}{18.02\ g/mol}\approx0.190\ mol\).

Step3: Determine reaction (assume \(2LiOH + CO_2 = Li_2CO_3 + H_2O\))

From reaction, moles of \(LiOH\) : moles of \(H_2O = 2:1\). Wait, no—wait, maybe the reaction is \(LiOH + H^+ = Li^+ + H_2O\) (simplified). Wait, actually, if we consider the formation of \(H_2O\) from \(LiOH\) reacting with an acid or \(CO_2\), but the key is the mole ratio. Wait, no—wait, the problem is to find moles of \(LiOH\) consumed to produce \(3.42\ g\) \(H_2O\). Let's re - check. Wait, maybe the reaction is \(2LiOH + H_2SO_4 = Li_2SO_4 + 2H_2O\), but no—wait, the standard reaction for \(LiOH\) and \(H_2O\) production: actually, the correct approach is: first, moles of \(H_2O\) is \(n=\frac{3.42}{18.02}\approx0.190\ mol\). Now, if the reaction is \(LiOH + H^+ = Li^+ + H_2O\), then moles of \(LiOH\) = moles of \(H_2O\). Wait, that can't be. Wait, maybe the reaction is \(2LiOH + CO_2 = Li_2CO_3 + H_2O\), so moles of \(LiOH\) is \(2\times\) moles of \(H_2O\). Wait, no, the problem's options include \(0.380\) which is \(2\times0.190\). Wait, I think I made a mistake earlier. Let's recalculate moles of \(H_2O\): \(n(H_2O)=\frac{3.42\ g}{18.02\ g/mol}\approx0.190\ mol\). Now, if the reaction is \(2LiOH + CO_2 = Li_2CO_3 + H_2O\), then moles of \(LiOH = 2\times\) moles of \(H_2O\). So \(n(LiOH)=2\times0.190\ mol = 0.380\ mol\). Wait, but the options have \(0.380\) and \(0.190\). Wait, maybe the reaction is \(LiOH + H_2O\) formation from \(LiOH\) and \(H^+\) (1:1 ratio). But that's conflicting. Wait, let's re - examine the problem. The problem says "how many moles of \(LiOH\) are consumed to produce this much \(H_2O\)". Let's assume the reaction is \(LiOH + H^+ = Li^+ + H_2O\) (1:1 mole ratio). Then moles of \(LiOH\) = moles of \(H_2O\approx0.190\ mol\). But if the reaction is \(2LiOH + CO_2 = Li_2CO_3 + H_2O\) (2:1 ratio), then moles of \(LiOH = 2\times0.190 = 0.380\ mol\). Wait, the options have both \(0.190\) and \(0.380\). Let's check the molar mass calculation again. Molar mass of \(H_2O\): \(H = 1.008\), so \(2\times1.008 = 2.016\), \(O = 16.00\), so total \(18.016\ g/mol\). Mass of \(H_2O\) is \(3.42\ g\), so moles \(=\frac{3.42}{18.016}\approx0.190\ mol\). Now, if the reaction is \(LiOH + H_2O\) - no, \(LiOH\) is a base, so when it reacts to form \(H_2O\), for example, with an acid: \(LiOH + HCl = LiCl + H_2O\). In this reaction, the mole ratio of \(LiOH\) to \(H_2O\) is \(1:1\). So moles of \(LiOH\) would be equal to moles of \(H_2O\), which is \(0.190\ mol\). But if the reaction is with \(CO_2\): \(2LiOH + CO_2 = Li_2CO_3 + H_2O\), then mole ratio \(LiOH:H_2O = 2:1\), so moles of \(LiOH = 2\times0.190 = 0.380\ mol\). The problem's options include both. Wait, maybe the intended reaction is the one with \(CO_2\), so moles of \(LiOH = 2\times\) moles of \(H_2O\). Wait, but let's check the options. The options are \(0.190\), \(0.380\), \(1.00\), \(18.02\). So if we consider the reaction \(2LiOH + CO_2 = Li_2CO_3 + H_2O\), then moles of \(LiOH = 2\times\frac{3.42}{18.02}\approx2\times0.190 = 0.380\ mol\). Wait, now I'm confused. Wait, maybe the problem is mis - stated, or the reaction is different. Wait, let's start over.

1. Calculate moles of \(H_2O\):
\(n(H_2O)=\frac{m(H_2O)}{M(H_2O)}=\frac{3.42\ g}{18.02\ g/mol}\approx0.190\ mol\).

2. Assume the reaction is \(LiOH + H^+ = Li^+ + H_2O\) (1:1 ratio). Then moles of \(LiOH = 0.190\ mol\).

3. If the reaction is \(2LiOH + CO_2 = Li_2CO_3 + H_2O\) (2:1 ratio), then moles of \(LiOH = 2\times0.190 = 0.380\ mol\).

Looking at the options, \(0.380\) is an option. Wait, maybe the correct reaction is the one with \(CO_2\), so the moles of \(LiOH\) is \(0.380\ mol\). Wait, but let's check the calculation again. Wait, \(3.42\ g\) of \(H_2O\), moles of \(H_2O=\frac{3.42}{18.02}\approx0.190\ mol\). If the reaction is \(2LiOH + CO_2 = Li_2CO_3 + H_2O\), then for every 1 mole of \(H_2O\) produced, 2 moles of \(LiOH\) are consumed. So moles of \(LiOH = 2\times0.190 = 0.380\ mol\). So the answer should be \(0.380\) moles. Wait, but the first step - by - step had a mistake in the reaction ratio. Let's correct it.

Step1: Calculate moles of \(H_2O\)

\(n(H_2O)=\frac{3.42\ g}{18.02\ g/mol}\approx0.190\ mol\).

Step2: Determine mole ratio (reaction \(2LiOH + CO_2 = Li_2CO_3 + H_2O\))

Mole ratio \(LiOH:H_2O = 2:1\).

Step3: Calculate moles of \(LiOH\)

\(n(LiOH)=2\times n(H_2O)=2\times0.190\ mol = 0.380\ mol\).

Answer:

\(0.380\) Moles