Be sure to answer all parts. Metal hydrides react with water to form hydrogen gas and metal hydroxide. For example, SrH2 (s) + 2H2O (l) → Sr (OH) 2 (s) + 2H2 (g) You wish to calculate the mass of hydrogen gas that can be prepared from 5.64 g of SrH2 and 4.70 g of H2O. (a) How many moles of H2 can be produced from the given mass of SrH2?

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Irene Randall

Chemistry

4 February, 09:14

Be sure to answer all parts. Metal hydrides react with water to form hydrogen gas and metal hydroxide. For example, SrH2 (s) + 2H2O (l) → Sr (OH) 2 (s) + 2H2 (g) You wish to calculate the mass of hydrogen gas that can be prepared from 5.64 g of SrH2 and 4.70 g of H2O. (a) How many moles of H2 can be produced from the given mass of SrH2?

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Jair Short · Answer 1

SrH₂ + 2H₂O = Sr (OH) ₂ + 2H₂

90gm 36gm 2 moles

5.64 g 4.7 g

water required for 5.64 g of SrH₂ = (36 / 90) x 5.64 g

= 2.256 g

water is in excess. Hence limiting reagent is SrH₂

90g SrH₂ makes 2 mole of water

5.64g SrH₂ makes water equal to mole = 2 x 5.64 / 90

=.125 mole.

mole of hydrogen formed =.125.