This experiment calls for the use of 2mL of 2-chloro-2methylpropane. The molar m

Question

This experiment calls for the use of 2mL of 2-chloro-2methylpropane. The molar mass of 2-chloro-2methylpropane is 92.57g mol-1, and its density is 0.847g mL-1. 1) Calculate the number of moles of 2-chloro-2methylpropane used in this experiment. 2) Calculate the number of moles of HCl produced by the complete reaction of 2 mL of 2-chloro-2-methylpropane. 3) Calculate the volume of 0.380M NaOH solution required to neutralize the total amount of HCl produced by the hydrolysis reaction. 4) Calculate the volume of 0.380M NaOH solution you will need to neutralize the HCl produced when the solvolysis reaction is 75% complete. 5) Will the buret contain sufficient NaOH solution to neutralize the HCl produced by the solvolysis reaction by the time the reaction is 75% complete?

Explanation / Answer

g mol^-1 means gram per mole since the density is .847 g ml^-1 and you are using 2ml, that means the mass of methylpropane used will be 1.694grams (2*0.847). now that you have mass you can find moles by taking mass and dividing by molar mass (1.694/92.57) and you get 0.0183 moles of methylpropane. since i don't know all the details of your problem i can't give you a definite answer to the next parts but if the mole ratio of your products to the reactants is a one to one ratio for every one mole of reactant you have you will get one mole of product out, therefore you have 0.0183 moles of methylpropane, you will get 0.0183 moles of HCl. since the molarity of the NaOH is 0.380 that means there are 0.380 moles of NaOH for every liter of solution, therefore 0.0183/0.380= 0.0482L of solution (or 48.2 mL) if the solvolysis reaction is 75% complete you should only get 0.0137 moles of HCl (0.0183 * 0.75) and then you will only need 0.0137/0.380=0.0361L of solution ( or 36.1mL)

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