Calcium hydride, CaH_2, reacts with water to form hydrogen gas: CaH_2 (s) + 2 H_
ID: 537370 • Letter: C
Question
Calcium hydride, CaH_2, reacts with water to form hydrogen gas: CaH_2 (s) + 2 H_2 O(l) rightarrow Ca(OH)_2 (aq) + 2 H_2 (g) This reaction is sometimes used to inflate life rafts, weather balloons, and the like, when a simple, compact means of generating H_2 is desired. How many grams of CaH_2 are needed to generate 145 L of H_2 gas if the pressure of H_2 is 825 torr at 21 degree C? The metabolic oxidation of glucose, C_6 H_12 O_4, in our bodies produces CO_2, which is expelled from our lungs as a gas: C_6 H_12 O_6 (aq) + 6 O_2 (g) rightarrow 6 CO_2 (g) + 6 H_2 O(l) (a) Calculate the volume of dry CO_2 produced at body temperature (37 degree C) and 0.970 atm when 24.5 g of glucose is consumed in this reaction. (b) Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 50.0 g of glucose. Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of Cl_2 gas is 8.70 L at 895 torr and 24 degree C. (a) How many grams of Cl_2 are in the sample? (b) What volume will the Cl_2 occupy at STP? (c) At what temperature will the volume be 15.00 L if the pressure is 8.76 times 10^2 torr? (d) At what pressure will the volume equal 500 L if the temperature is 58 degree C?Explanation / Answer
we know that 825 torr = 1.085 atm pressure
one mole of all gases occupies 22.414 L volume at one atm pressure and at 273 K
As per Charles's law,
V1/T1 = V2/T2
using this we can find the volume of one-mole of gas at 21 oC or 294 K
(22.414/273)*294 = V2 = 24.13 L
Now we have to find the volume of one mole of gas at 1.085 atm
By applying Boyle's law
P1V1 = P2V2
1*24.13 = 1.085 * V2
V2 = 22.23 L
145 L hydrogen contains = 145/22.23 = 6.52 moles
As per the balanced equation, one mole C2H2 produce one-mole hydrogen gas
Thus for 6.52 moles 6.52 c2H2 is used
molecular weight of C2H2 is 26 g/mole
thus weight of C2H2 is = 26 x 6.52 = 169.59 g
As per the policy of Chegg one question is answered.