Calcium hydride, CaH2, reacts with water to form hydrogen gas: CaH2(s)+2H2O(l)Ca
ID: 958295 • Letter: C
Question
Calcium hydride, CaH2, reacts with water to form hydrogen gas: CaH2(s)+2H2O(l)Ca(OH)2(aq)+2H2(g) This reaction is sometimes used to inflate life rafts, weather balloons, and the like, where a simple, compact means of generating H2 is desired.
Part A
How many grams of CaH2 are needed to generate 146 L of H2 gas if the pressure of H2 is 824 torr at 22 C?
The metabolic oxidation of glucose, C6H12O6, in our bodies produces CO2, which is expelled from our lungs as a gas: C6H12O6(aq)+6O2(g)6CO2(g)+6H2O(l)
Part A
Calculate the volume of dry CO2 produced at body temperature (37 C) and 0.990 atm when 23.5 g of glucose is consumed in this reaction
Part B
Calculate the volume of oxygen you would need, at 1.00 atm and 298 K, to completely oxidize 53 g of glucose.
Explanation / Answer
Answer – Part A ) Given, volume of H2 gas = 146 L , pressure , P = 824 torr, T = 22 +273 = 295 K
Reaction - CaH2(s)+2H2O(l) ----> Ca(OH)2(aq)+2H2(g)
Calculation of moles of H2 using Ideal gas law
Convert the P torr to atm
We know,
760 torr= 1 atm
So, 824 torr = ?
= 1.08 atm
We know Ideal gas law
PV = nRT
So, n = PV /RT
= 1.08 atm * 146 L / 0.0821 L.atm.mol-1.K-1*295 K
= 6.54 moles of H2
From the balanced equation –
2 moles of H2 = 1 moles of CaH2
So, 6.54 moles of H2 = ?
= 3.27 moles of CaH2
Moles of CaH2 to mass –
Mass of CaH2 = moles of CaH2 * molar mass of CaH2
= 3.27 moles * 42.094 g/mol
= 137.6 g of CaH2
137.6 grams of CaH2 are needed to generate
Part A) We are given, P of CO2 = 0.990 atm , T = 37 +273 = 310 K
Mass of glucose = 23.5 g
Reaction - C6H12O6(aq)+6O2(g) -----> 6CO2(g)+6H2O(l)
Calculation of moles of glucose –
moles of glucose = 23.5 g / 180.15 g.mol-1
= 0.130 moles
Calculation of moles of CO2
From the balanced equation –
1 moles of C6H12O6(aq) = 6 moles of CO2
So, 0.130 moles of C6H12O6(aq) = ?
= 0.783 moles of CO2
Now using the Ideal gas law –
PV = nRT
So, V = nRT/P
= 0.783 moles * 0.0821 L.atm.mol-1.K-1*310 K / 0.990 atm
= 20.1 L
the volume of dry CO2 produced at body temperature is 20.1 L
Part B) Given, mass of glucose = 53 g, P = 1.0 atm, T = 298 K
Calculation of moles of glucose –
moles of glucose = 53 g / 180.15 g.mol-1
= 0.294 moles
Calculation of moles of O2
From the balanced equation –
1 moles of C6H12O6(aq) = 6 moles of O2
So, 0.294 moles of C6H12O6(aq) = ?
= 1.76 moles of CO2
Now using the Ideal gas law –
PV = nRT
So, V = nRT/P
= 1.76 moles * 0.0821 L.atm.mol-1.K-1*298 K / 1.00 atm
= 43.2 L
volume of oxygen is 43.2 L