lostinIB 0 Posted October 18, 2019 Report Share Posted October 18, 2019 Hi there, I'm really struggling with topic 1 stoichiometry and I really need help! Will you please help me with the following 3 questions? 1. 5.0dm3 of 2.00mol dm-3 sodium carbonate solution, Na2CO3 (aq)was added to a volumetric flask and the volume was made up to 500 cm3 with water. What is the concentration in mol dm-3, of the solution? 2. How many grams of sodium azide, are needed to produce 68.1dm3 of N2 (g) at STP? Molar V at STP = 22.7 dm 3 mol-1 ; Mr(NaN3) = 65.0 2NaN3 (s) → 3N2 (g) + 2Na (s) 3. 2.478g of white phosphorous was used to make phosphine according to the equation: P4 +3OH- +3H2O ---- PH3 +3H2PO a) Calculate amount in mol of white phosphorus used b) This phosphorus was reacted with 100.cm3 of 5.00 moldm-3 of aqueous sodium hydroxide. Deduce, showing your working, which was the limiting reagent. c) Determine excess amount in mol of the other reagent d) Determine the volume of phosphine, measured in cm2 at STP that was produced. For #3, I'm stuck after the first a), and I have no clue how to start off with number 1 and 2. Is #1 a question related to titration or concentration? For question 2, do I use the volume of gas law with the volume ratio to solve the question? Thank you, please help me!! Reply Link to post Share on other sites

benjeren 2 Posted November 2, 2019 Report Share Posted November 2, 2019 Q1 I use the formula: concentration = moles / volume (C = n/V) We are given "5.0dm3 of 2.00mol dm-3 sodium carbonate solution". What I did was rearrange the question to find the moles of the solution so I did: n1 = C1V1 (i added 1's to make it look clear in the next step) Therefore: C = C1V1 / V (=) C1V1 = 5.0dm3 x 2.00 = 10 mol/dm3 (=) C = 10/ (500/1000)dm3 = 20 mol/dm3 Q2 PV = nRT --> n = RT/PV STP values are: T=273K R=8.31 (forgot what the units for this was) P=1.00x10ˆ5 Pa Substitute: n = (1.00x10ˆ5Pa)(0.0618m3) / (8.31)(273K) = 3.002 mol of N2 The molar ratio between sodium azide and nitrogen is 2:3 Therefore --> (2/3)(3.002) = 2.001 moles of sodium azide Using the formula: moles = mass / molar mass (n=m/M) rearranged to m=nM: 2.001 mol X 65.0 g/mol = 130.0g Q3 a) n=m/M M = 30.97 g/mol 2.478g / 30.97 g/mol = 0.08 mol. of phosphorus b) C=n/V to n=CV we can get moles of aqueous sodium hydroxide: (100/1000)dm3 x 5 mol/dm3 = 0.5 mol of sodium hydroxide We divide both moles of solution by their coefficients and the smaller value is limiting: 0.08/1 = 0.08 (limiting) so the phosphorous is limiting 0.5/3 = 1.67 c) I'm not sure about this question but this is how i interpreted it: (100/1000)dm3 x 5 mol/dm3 = 0.5 mol of sodium hydroxide Molar Ratio is 1 mol. of Phosphorous: 3 mol. of sodium hydroxide 0.5/3 = 1.67 d) I think you meant cm3 instead of cm2: moles x 22.7 (STP volume) = volume 0.5 x 22.7 = 11.35 cm3 NOTE: I'm not sure about the last two parts of question 3 but I can ask my teacher. I've got a terrible teacher so I have to self-teach myself chemistry at the moment so I'm sorry if I got anything wrong here/did not match the markscheme 1 Reply Link to post Share on other sites

kw0573 1,254 Posted December 26, 2019 Report Share Posted December 26, 2019 @benjeren 1. There was a typo in the original question as you can't concentrate a solution by adding more water. Likely it was 5.0 cm3 of initial solution, at which the new concentration is just 100 fold more dilute or 0.010 mol dm-3 2. I concur with your answer, but clearly you meant n = PV/RT. 3. a) The phosphorus occurs in P4 in its elemental form, as provided in the question. So the molar mass is 4 × 30.974 = 123.90 g/mol 2.478 g / (123.90 g/mol) = 0.0200 mol of P4. b) P4 is still limiting, but 0.5/3 is 0.167 c) 0.500 mol of hydroxide - 3mol of OH/mol of P4 * (0.020 mol of P4) = 0.440 mol of hydroxide in excess d) phosphine is PH3, which is in 1-to-1 mol ratio as P4. Also 22.7 is in dm3, so you need to use 22 700 cm3 0.02 mol (22 700 cm3/mol) = 454 cm3. Reply Link to post Share on other sites

## Recommended Posts

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.