Question

How much calcium metal must be added to an excess of water to produce 3.7g of calcium hydroxide?

Answer 1

The mass of calcium is `=2.00g`

Explanation:

THe reaction of calcium with water is

`"calcium" + "water"rarr "calcium hydroxide" + "hydrogen"`

`Ca(s)+ H2O rarr Ca(OH)_2 + H_2(g)`

`1` mole of Calcium is `=40.078g`

`1` mole of Calcium hydroxide is

`=40.078+(2*15.999)+(2*1.008)=74.092g`

To produce `3.7g` of calcium hydroxide,

The mass of calcium needed is `=40.078*3.7/74.092=2.00g`

Answer 2

About `2*g`...

Explanation:

As always we need a stoichiometric equation to inform our reasoning:

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

And we work out (i) the moles of calcium hydroxide...

`"Moles of calcium hydroxide"-=(3.70*g)/(74.09*g*mol^-1)=0.0499*mol`...

And so (ii) we need an equivalent molar quantity of metal....

i.e. `0.0499*molxx40.1*g*mol^-1=2.00*g`