Limiting reactant problem, chemistry?

Question

Can you please check my answers?

 

What is the maximum mass of hydrogen gas that can be produced from 5.70 g SrH2 and 4.75g H2O? 

 

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

 

Assuming that all molar mass calculations will be based on the atomic mass rounded to two decimal places, I got the following values: 

 

5.70 g SrH2 will produce 0.257 g H2, and 4.75 g H2O will produce 1.06 g H2. Thus, the maximum mass of hydrogen gas that can be produced is 1.06 g. 

 

If I am way off, please let me know. Thank you for your time!

 

Answer 1

What is the maximum mass of hydrogen gas that can be produced from 5.70 g SrH2 and 4.75g H2O?

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

 

Limiting reactants are based on MOLES present (not grams) and the mole ratio of the reactants.

 

molar mass SrH2 = 89.64 g

molar mass H2O = 18.02 g

moles SrH2 present = 5.70 g x 1 mol/89.64 g = 0.0636 moles

moles H2O present = 4.75 g x 1 mol/18.02 g = 0.2636 moles

moles H2O needed for 0.0636 moles SrH2 = 2 x 0.0636 = 0.1272 moles, so H2O is NOT limiting, but SrH2 is limiting

moles H2 produced from 0.0636 moles SrH2 = 2 x 0.0636 = 0.1272 moles (see 2:1 stoichiometry for H2: SrH2)

mass of H2 produced = 0.1272 moles x 2 g/mol = 0.254 g

Answer 2

The approach to limiting reactant problems with two reactants is to compute the mass of product formed two ways:

    1) by assuming that the first reactant is limiting and

    2)  by assuming that the second reactant is limiting.

 

 

   The actual limiting reactant is the one corresponding to the lesser of 1) and 2).

 

  So, for the problem at hand, SrH₂  is the limiting reactant and the maximum H₂ that can be produced is 0.257g.