How many litres of hydrogen gas can be generated by reacting 6.25 g of barium hydride with water at 20oC and 755 mm Hg pressure, according to the following chemical equation?

BaH2(s) + 2H2O(l)? Ba(OH)2(aq) + 2H2(g)

Question

How many litres of hydrogen gas can be generated by reacting 6 25 g
of barium hydride with water at 20oC and 755 mm Hg pressure,
according to the following chemical equation?

BaH2(s) + 2H2O(l)? Ba(OH)2(aq) + 2H2(g)

Kushal Kumar · Accepted Answer

Dear Student
Volume of gas 1 mol H2 at 0 degree celsius and 760 mm Hg= 22
4 LitresSo volume of H2 at 20 degree celsius and 755 mm Hg= V2Now applying P1V1/T1 = P2V2/T2Here P1= 760 mm HgP2 = 755 mm HgV1=22
4 litresT1=273 KT2 = 293 KSubstituting the values and doing calculations we get V2 as 24
2 litresnow moles of BaH2 = 6
25/139=0
044So moles of H2 produced = 0
088Hence volume produced = 24 2 x 0
088=2 1296 litres
Regards

How many litres of hydrogen gas can be generated by reacting 6.25 g of barium hydride with water at 20oC and 755 mm Hg pressure, according to the following chemical equation? BaH2(s) + 2H2O(l)? Ba(OH)2(aq) + 2H2(g)

How many litres of hydrogen gas can be generated by reacting 6.25 g of barium hydride with water at 20oC and 755 mm Hg pressure, according to the following chemical equation?

BaH2(s) + 2H2O(l)? Ba(OH)2(aq) + 2H2(g)