Respuesta :
NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)
You want to calculate the mass of NaH, I assume. Otherwise, the question isn't clear. It simply says calculate the mass(??)
So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr. But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas. So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT
n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)
n = 0.0385 moles H2
moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required
mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
Brainliest Please :)
0.925g moles NaH weighing to produce 982 mL of gas at 28.0 °C and 765 torr, when the hydrogen is collected over water.
What is displacement reaction?
Displacement reaction is a chemical reaction in which are more reactive element displaces a less reactive element from its compound both metals and nonmetals take a part in displacement reaction
So, calculate the moles of H₂ gas that satisfy the conditions of 982 ml at 28ºC and 765 torr.
But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H₂ gas.
So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.
PV = nRT (Ideal gas equation)
n = PV/RT = (737 torr)x(0.982 L)/(62.4 L-torr / K mol)x(301 K)
n = 0.0385 moles H₂
Moles NaH needed = 0.0385 moles H₂x 1 mole NaH/mole H2 = 0.0385 moles NaH required
Mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH
0.925g moles NaH weighing to produce 982 mL of gas at 28.0 °C and 765 torr, when the hydrogen is collected over water.
Learn more about displacement reaction, Here:
https://brainly.com/question/3172917
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