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?

Explanation:

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 .

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml