When strontium hydride reacts violently with water to produce strontium hydroxid

Question

When strontium hydride reacts violently with water to produce strontium hydroxide and hydrogen gas. The unbalanced reaction is shown below. Balance the chemical reaction. We wish to calculate the maximum mass of hydrogen gas that can be produced when 5.63 grams of SrH_2 is reacted with 4.80 grams of H_2 O. SrH_2 (s) + H_2 O (l) rightarrow Sr(OH)_2 (s) + H_2 (g) Calculate how many moles of H_2 can be formed given the mass of SrH_2. Calculate how many moles of H_2 can be formed given the mass of H_2 O Which of the two reactants is the limiting reactant? How many grams of H_2 can be formed?

Explanation / Answer

Balanced chemical reaction:

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

a) 1 mole of SrH2 produce 2 moles of H2

Now,mass of SrH2 = 5.63 g

Moles = mass/molar mass

Molar mass of SrH2 = 89.6g/mol

So,moles of SrH2 = 5.63/89.6 = 0.063 moles

I mole of SrH2 produce 2 moles of H2

So 0.063 moles of SrH2 produce 0.126 moles of H2 gas.

b). From the reaction,

2 moles of H2O produce 2 moles of H2

So 1 mole of H2O produce 1 mole of H2

Mass of H2O = 4.80g

Molar mass of H2O = 18g/mol

Moles of H2O = 4.80/18 = 0.267moles

So 0.267 moles of H2O produce 0.267 moles of H2 gas.

c).So,0.063 moles of SrH2 reacts with 0.267 moles of H2O

1 mol-rxn = 1moleSrH2 = 2molesH2O

0.063mol of SrH2 × 1mol-rxn/1 mol SrH2 = 0.063mol-rxn

0.267mol of H2O × 1 mol-rxn/2 mol H2O = 0.133mol -rxn

The reactant with fewer mole-reaction is the limiting reagent

Hence,SrH2 is the limiting reagent here.

d).Since SrH2 is the limiting reagent.

Which produces 0.126 moles of H2 gas

Mass of H2 produced = moles × molar mass = 0.126 × 2 = 0.252g of H2.

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