What mass of calcium when reacted with water will produce 15.7 L of hydrogen at STP?

Now it is a fact that 1 mol of gas has a volume of #22.4* dm^3# at STP assuming ideal behaviour. According to the reaction above, 1 mol of calcium metal (#40.08*g#) should produce #22.4*L# gas.

By this stoichiometry, #(15.7*Lxx40.08*g *mol^(-1))/(22.4*L*mol^-1)# #Ca# were used (about 32 g?). (NB #1*L=1*dm^3#)

Balanced Equation

#"Ca(s)" + 2"H"_2"O"##rarr##"Ca(OH)"_2("s") + "H"_2("g")"#

#"STP"# is #"273.15 K"# and #"100 kPa"#.

The molar volume of a gas at #"273.15 K"# and #"100 kPa"# is #"22.710 mol/L"#

Molar mass of #"Ca = 40.078 g/mol"# (atomic weight in g/mol)

The process:

#color(red)("L H"_2")##rarr##color(red)("mol H"_2")##rarr##color(green)("mol Ca")##rarr##color(blue)("mass Ca")#

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#color(red)("L H"_2")##rarr##color(red)("mol H"_2")#

First determine moles #"H"_2"# by dividing the given volume by the molar volume #("22.710 L/mol")#. I prefer to divide by multiplying by the reciprocal of its molar volume #("1 mol/22.710 L")#.

#15.7color(red)cancel(color(black)("L H"_2))xx(1"mol H"_2)/(22.710color(red)cancel(color(black)("L")))="0.6913 mol H"_2"#

#color(red)("mol H"_2")##rarr##color(green)("mol Ca")#

To get mol #"Ca"#, multiply mol #"H"_2"# by the mole ratio between #"Ca"# and #"H"_2"# from the balanced equation, with #"Ca"# in the numerator.

#0.6913color(red)cancel(color(black)("mol H"_2))xx(1"mol Ca")/(1color(red)cancel(color(black)("mol H")))="0.6913 mol Ca"#

#color(green)("mol Ca")##rarr##color(blue)("mass Ca")#

To determine mass of #"Ca"# multiply times its molar mass.

#0.6913color(red)cancel(color(black)("mol Ca"))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))="27.7 g Ca"# (rounded to three significant figures)

You can put all three steps together into one equation.

#15.7color(red)cancel(color(black)("L H"_2))xx(1color(red)cancel(color(black)("mol H"_2)))/(22.710color(red)cancel(color(black)("L H"_2)))xx(1color(red)cancel(color(black)("mol Ca")))/(1color(red)cancel(color(black)("mol H"_2)))xx(40.078"g Ca")/(1color(red)cancel(color(black)("mol Ca")))="27.7 g Ca"# (rounded to three significant figures)

Note: If your teacher is still using STP as #0^@"C"# and #"1 atm"#, substitute #"22.414 L/mol"# for #"22.710 L/mol"#.

The mass of #"Ca"# would be #"28.1 g Ca"#.