Respuesta :
Answer:
[tex]m_{CaH_2}=37.4gCaH_2[/tex]
Explanation:
Hello,
In this case, since the undergoing chemical reaction is:
[tex]CaH_2(s) + 2 H_2O(l) \rightarrow Ca(OH)_2(aq) + 2 H_2(g)[/tex]
By knowing the gas data of hydrogen, we can compute the formed moles by considering the ideal gas equation as:
[tex]n_{H_2}=\frac{PV}{RT}=\frac{0.811atm*55.0L}{0.082\frac{atm*L}{mol*K}*(32+273.15)K} =1.78molH_2[/tex]
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:
[tex]m_{CaH_2}=1.78molH_2*\frac{1molCaH_2}{2molH_2}*\frac{42gCaH_2}{1molCaH_2} \\m_{CaH_2}=37.4gCaH_2[/tex]
Best regards.
Answer:
m = 37.5373 g
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
