Calcium hydride combines with water according to the equation: CaH_2(s) + 2H_2 O

Question

Calcium hydride combines with water according to the equation: CaH_2(s) + 2H_2 O(l) rightarrow 2H_2(g) + Ca(OH)_2(s) Beginning with 84.0 g of CaH_2 and 42.0 g of H_2 O, what volume of H_2 will be produced at 273 K and a pressure of 1327 torr? 29.9 L 15.0 L 5.39*10^2 L 25.7 L none of these Oxygen gas, generated by the reaction 2KClO_3(s) rightarrow 2KCl(s) + 3O_2(g) is collected over water at 27 degree C in a 1.55-L vessel at a total pressure of 1.00 atm.(The vapor pressure of H_2 O at 27 degree C is 26.0 torr.) How many moles of KClO_3 were consumed in the reaction? 0.0608 moles 0.0912 moles 0.0405 moles 0.0434 moles 1.50 moles

Explanation / Answer

28.

CaH2 + 2H2O --- > 2H2 + Ca(OH)2

Moles of CaH2 = 84.0g / 42.09388 g/ mol

= 1.99 or 2 mol

Moles of H2O = 42.0 g /18.02 g/mol

= 2.33 mol

Here H2O is limiting agent

Moles of H2:

2.33 mol H2O * 2 mol H2/ 2.0 mol H2O

= 2.33 mol H2

Here P = 1372 torr = 1.81 atm , T = 273 K and R = 0.0821 L atm / K.mol

PV = nRT (ideal gas law)

V= nRT/P

V= 2.33* 0.0821 *273/ 1.81

V= 28.9 L

Hence the correct answer is E.

29

2KClO3 --- > 2KCl +3O2

now calculate the mole sof O2:

PV= nRT

n = PV/RT

n= 0.966 *1.55/ 0.0821*300

n= 0.0608 mol

Now calculate the moles of KClo3

0.0608 mol O2 / 2 mol of KClO3 / 3 mol O2

= 0.0405 mol KClO3.

Correct answer is C

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---