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Sagot :
Answer:
3.73 g of Li.
Explanation:
What is given?
Volume of hydrogen gas, H2 (V) = 6.00 L.
Pressure (P) = 1.10 atm.
Temperature (T) = 25.0°C + 273 = 298 K.
R = 0.082 L*atm/mol*K.
Step-by-step solution:
First, we have to find the number of moles of hydrogen gas (H2) using the ideal gas formula:
[tex]PV=nRT,[/tex]where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant and T is the temperature on Kelvin scale. So let's solve for 'n' and replace the given values:
[tex]n=\frac{PV}{RT}=\frac{1.10\text{ atm}\cdot6.00\text{ L}}{0.082\text{ }\frac{L\cdot atm}{mol\cdot K}\cdot298K}=0.270\text{ moles.}[/tex]Based on this, we have 0.270 moles of H2.
You can see in the chemical equation that 2 moles of Li reacted produce 1 mol of H2, so we have to state a rule of three:
[tex]\begin{gathered} 2\text{ moles Li}\rightarrow1\text{ mol H}_2 \\ ?\text{ moles Li}\rightarrow0.270\text{ moles H}_2 \end{gathered}[/tex]The calculation will look like this:
[tex]0.270\text{ moles H}_2\cdot\frac{2\text{ moles Li}}{1\text{ mol H}_2}=0.540\text{ moles Li.}[/tex]The final step is to convert 0.540 moles of Li to grams using its molar mass that can be found in the periodic table, which is 6.9 g/mol:
[tex]0.540\text{ moles Li}\cdot\frac{6.9\text{ g Li}}{1\text{ mol Li}}=3.73\text{ g Li.}[/tex]The answer is that we must use 3.73 g of Li to produce 6.00 L of H2 at 1.10 atm and 25.0 °C.
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