Nt1310 Unit 1 Lab Report

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Okay, if our lithium weight is going to be 6.941 g/moL Then that means we have to take 24.6g of Lithium and multiply it by 1 mol of Lithium over 6.941 g of Lithium. This would equal to be 3.544 mol of Lithium. Then we have to take that 3.544 and multiply it by 1 mol of hydrogen gas over 2 mol of lithium. Which would then equal into 1.772 mol of hydrogen gas. We can then figure out that 1.772 is our “n”. The “T” is our 301 Kelvin, the “P” is our 1.01 atm and the “R” is our 0.0820 which would be the L atm over mol k. And we can’t forget about our “V” which would be V equals nRT over P which equals 1.772 mol divided by 0.0820 L atm over mol kelvin multiplied by 301 kelvin over 1.01 atm which equals to our final answer of: 43.33 of H2 If a mol

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