Question

a sample of cesium metal reacted completely with water, evolving 48.1 ml of dry H2 at 19C and 768 mmHg. what is the equation for the reaction? what was the mass of cesium in the sample ?

Answer 1

the reaction for equation is written as follows

2Cs (s) + 2H2O(l) = 2CsOH (aq) + H2 (g)

The mass of cesium is calculated as follows

Find the moles of H2 by use of ideal gas equation that is Pv = nRT
where n is the number of moles

n = PV/RT
P= 48.1 ml in liters = 48.1 /1000= 0.0481 l
T= 19 + 273.15 = 292.15K
P= 768mm hg
R (gas Constant)= 62.364 l.mmhg/k.mol

n= ( 768mmhg x0.0481 L) /( 62.364 L.mm hg/k.mol x 292.15k) = 2.028 x10^-3 moles

by use of mole ratio from reacting equation between Cs to H2 which is 2 :1 the moles of Cs is therefore = ( 2.028 x10^-3) x 2 = 4.05 x10^-3 moles

mass of cs is therefore = moles x molar mass of cs( 132.9g/mol)

=( 4.05 x10^-3)mol x 132.9 g/mol = 0.539 grams