Question

Calcium hydride (CaH2) reacts with water to form hydrogen gas: CaH2(s) + 2 H2O(l) → Ca(OH)2(aq) + 2 H2(g) Determine the number of grams of CaH2 are needed to generate 55.0 L of H2 gas at a pressure of 0.811 atm and a temperature of 32°C.\

Answer 1

m = 37.5373 g

Step-by-step explanation:

To do this, let's write the reaction again:

CaH₂(s) + 2H₂O(l) --------> Ca(OH)₂(aq) + 2H₂(g)

In this reaction we can see that the mole ratio of CaH₂ and H₂ is 1:2, in other words:

moles CaH₂/moles H₂ = 1/2

We need to know how many grams of CaH₂ reacted to form 55 L of Hydrogen. to do this, we need to know how many moles of Hydrogen were produced. and as Hydrogen is a gas, and we know the volume and pressure of it, we can use the ideal gas equation to solve for the moles of hydrogen:

PV = nRT solvinf for the moles:

n = PV / RT

R is constant of gases and is 0.082 L atm /mol K and the temperature is 32 °C (305 K) so, the moles are:

n = 0.811 * 55 / 0.082 * 305

n = 1.7835 moles

We have the moles, let's calculate the moles of CaH₂ used with the mole ratio of before:

moles CaH₂/moles H₂ = 1/2

moles CaH₂ = moles H₂/2

moles CaH₂ = 1.78 / 2 = 0.89175 moles

Finally the mass is calculated using the molecular mass of CaH₂ which is 42.094 g/mol, so the mass is:

m = 0.89175 * 42.094

m CaH₂ = 37.5373 g

Answer 2

`m_(CaH_2)=37.4gCaH_2`

Step-by-step explanation:

Hello,

In this case, since the undergoing chemical reaction is:

`CaH_2(s) + 2 H_2O(l) \rightarrow Ca(OH)_2(aq) + 2 H_2(g)`

By knowing the gas data of hydrogen, we can compute the formed moles by considering the ideal gas equation as:

`n_(H_2)=(PV)/(RT)=(0.811atm*55.0L)/(0.082(atm*L)/(mol*K)*(32+273.15)K) =1.78molH_2`

In such a way, since hydrogen and calcium hydride have a 2 to 1 molar relationship in the chemical reaction, the number of grams of calcium hydride turns out, by means of stoichiometry:

`m_(CaH_2)=1.78molH_2*(1molCaH_2)/(2molH_2)*(42gCaH_2)/(1molCaH_2) \\m_(CaH_2)=37.4gCaH_2`

Best regards.