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

2 Answers
Jan 30, 2018

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

Jan 30, 2018

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