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Question

A student reacted with the following unknown metal, M, with HCI and collecyed the following data : mass of metal = 0.300, volume of water displaced = 227.3 mL, temperature of H2 =25C , barometric pressure = 760.0 torr, volume of HCI=20.0 ml, molarity of HCI = 1.500 mol/L and volume of NaOh =36.0 mL. If the metal, M, had an atomic weight of 50.0 amd reacted to produce MCI3, what was the molarity of the NaOH solution?

Answer 1

Answer is : molarity of NaOH is 0.33M

 

Solution is shown stepwise.

1) mol of metal: mass/molar mass = 0.3(g)/50(g/mol) = 0.006 mol or 6 mmol of metal

2) mol of HCl: molarity x volume = 0.020L x 1.5M = 0.03 mol or 30 mmol of HCl

3) balanced reaction between metal and acid: 2M + 6 HCl = 2MCl3 + 3H2

4) since metal is limiting reagent, then acid is excess reagent, hence via stoichiometry, we know that we will have at the end of the reaction: 0 mmol of metal (all consumed) and 12 mmol of HCl (6x6/2 = 18 mmol consumed).

5) to check our result of step 4, we do PV=nRT, where n = RT/(PV), and we indeed find that n = 9 mmol of H2 gas, which corresponds to the equation we wrote at Step 3 (6x3/2 = 9 mmol produced). This confirms that Steps 3-4-5 are correct.

6) The excess HCl at the end of the reaction (12 mmol HCl) are then neutralized by the NaOH. We need 12 mmol of NaOH to neutralize 12 mmol of acid.

7) molarity NaOH = mol/volume = 12 (mmol)/36 (ml) = 0.33 M

 

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