Question

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

Answer 1

`Ca(s) + 2H_2O(l) rarr Ca(OH)_2(aq) + H_2(g)uarr`

Explanation:

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`)

Answer 2

`"27.7 g Ca"` are required to produce `"15.7 L H"_2"`.

Explanation:

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"`.