Question

Metal hydrides react with water to form hydrogen gas and the metal hydroxide. Use significant figures.

SrH2(s) + 2 H2O(l) Sr(OH)2(s) + 2 H2(g)

You wish to calculate the mass of hydrogen gas that can be prepared from 4.71 g of SrH2 and 4.26 g of H2O.
(a) How many moles of H2 can be produced from the given mass of SrH2?
(b) How many moles of H2 can be produced from the given mass of H2O?
(c) Which is the limiting reactant? (Type your answer using the format CO2 for CO2.)
(d) How many grams of H2 can be produced?

Answer 1

molar mass of SrH2 = 89.6 g/mol
molar mass of H2O = 18 g/mol
number of mol of SrH2 = (given mass)/(molar mass)
= 4.71/89.6
= 0.0526 mol
number of mol of H2O = (given mass)/(molar mass)
= 4.26/18
= 0.237 mol

reaction taking place is
SrH2(s) + 2 H2O(l) Sr(OH)2(s) + 2 H2(g)

according to reaction
1 mol of SrH2 react with 2 mol of H2O
0.0526 mol of SrH2 react with (2*0.0526) mol of H2O
0.0526 mol of SrH2 react with 0.11 mol of H2O
but we have 0.237 mol of H2O
so , H2O is in excess
and hence, SrH2 is limiting reagent

a)
1 mol of SrH2 give 2 mol of H2
0.0526 mol of SrH2 give (2*0.0526) mol of H2
number of mol of H2 produce = (2*0.0526) mol
= 0.11 mol

Answer : 0.11 mol
b)
Answer : 0.11 mol
because once SrH2 get over no further reaction take place
c)
SrH2
d)
molar mass of H2 = 2 g/mol
mass of H2 produce = (number of mol of H2)*(molar mass of H2)
= (0.11*2) g
= 0.22 g

Answer : 0.22 g