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.

Answer 1

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