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# TRIPLE Calculations

### TRIPLE AQA GCSE chapter 04 calculations

In an experiment 2.4g of Magnesium is burned; and 3.6g of Magnesium Oxide is collected. What is the yield of Magnesium Oxide? 3.6g (what you should collect)
The theoretical yield of Magnesium Oxide is 4.0g. But only 3.6g is collected. What is the %yield? 3.6/4.0*100=90%
A factory makes 100 moles of ammonia gas. What volume would this gas occupy at room temperature and pressure? 1mole occupies 24dm3 so 100 moles would occupy 2400dm3
3H2(g) + N2(g) ïƒ³ 2NH3(g) What is the minimum volume of hydrogen needed to react completely with 300 cm3 of nitrogen? because it is gas to gas reaction, we simply need to look at the ratio 3 H2:1 N2 so 3 times as much Hydrogen ie 900cm3
1 mol of magnesium, Mg, reacts to give 1 mol of hydrogen, H2. The experiment shows that 0.072 g of magnesium reacts to produce 66 cm3 of hydrogen. Calculate the volume of 1 mol of hydrogen, H2 (relative atomic mass: Mg = 24) this is a mass to volume so sucue; moles Mg = 0.072/24=0.0030; moles H2=moles Mg; volume =66/0.003=22 000 cm3; note that HERE we can NOT assume that volume is 24000cm3
When the experiment was repeated (Mg placed in acid), the volume of 1 mol of hydrogen, H2, was found to be 23000 cm3. The molar volume of hydrogen is 24000 cm3 at room temperature and atmospheric pressure. Suggest why these volumes are different temperature lower/pressure higher/ some hydrogen escapes (1) hence calculated volume lower (1)
Calcium reacts with water to produce hydrogen. Ca + 2H2O â†’ Ca(OH)2 + H2 Calculate the volume of H2, that could be produced from 15 g of calcium. ( Ca = 40, 1 mol of hydrogen, H2, occupies 24000 cm3 at room temperature and atmospheric pressure) this is a mass to volume so sucue; moles Ca = 15/40= 0.375; molesH2=moles Ca;volume H2 = 15 x 24000(1) = 9000 cm3
N2(g) + 3H2(g) ïƒ³ 2NH3(g). Calculate the minimum volume of hydrogen required to completely convert 1000 dm3 of nitrogen into ammonia. because it is gas to gas reaction, we simply need to look at the ratio 3 H2:1 N2 so 3 times as much Hydrogen ie 3000dm3
Ammonia is reacted with excess nitric acid, HNO3, to make ammonium nitrate, NH4NO3. NH3 + HNO3 â†’ NH4NO3 Calculate the mass of ammonium nitrate produced by the reaction of 34 g of ammonia. (Relative atomic masses H = 1.0, N = 14, O = 16) this is a mass to volume so sucue; RFM (NH4NO3)=80; RFM(NH3)=17; moles NH3=34/17=2; moles NH4NO3 = moles NH3; mass NH4NO3=molesxRFM=2x80=160g
Hydrogen reacts with oxygen to form water vapour. 2H2(g) + O2(g) <=> 2H2O(g) If 200 cm3 of H2 react completely with 100 cm3 of oxygen, what is the maximum volume of H2O vapour formed, if all volumes are measured at the same temperature and pressure? because it is gas to gas reaction, we simply need to look at the ratio 2 H2:2 H2O so same volume of H2 and H2O ie 200dm3
Zn(s) + 2HCl(aq) ïƒ  ZnCl2(aq) + H2(g) Calculate the volume of hydrogen formed, when 13.0 g of zinc reacts completely with excess hydrochloric acid. (relative atomic mass: Zn = 65.0, 1 mol of any gas occupies 24 dm3 at room temperature and pressure) this is a mass to volume so sucue;moles Zn=13/65=0.2; moles H2=moles Zn; vol H2=molesx24=4.8dm3
N2(g) + 3H2(g) ïƒ³ 2NH3(g) Calculate the minimum volume of nitrogen, in dm3, required to react completely with 1000 dm3 of hydrogen. All volumes are measured at the same temperature and pressure. simply look at the ratio because it is gas to gas reaction, 3 H2:1 N2 so 3 times as much Hydrogen ie 333dm3 of Nitrogen needed
What is the unit of concentration? mol.dm-3
What is the amount(moles) of solution in 24.35cm3 of a 0.125Mole.dm-3 solution of Potassium Nitrate KNO3? moles = 24.35/1000 x 0.125 = 0.00304 or 3.04x10^-3 moles
What is the concentration of 8.75x10^-3 moles in 50cm3 of solution? Concentration = 8.75x10^-3 / (50/1000)= 0.175 mole.dm-3 Have you remembered to convert cm3 into dm3? Have you remembered the units?
Created by: ursulinechem2