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

Question

Wyatt Atkins · Accepted Answer

About #2*g#...

As usual, our reasoning must be guided by the following stoichiometric equation: #Ca(s) + 2H_2O(l) rarr Ca(OH)_2(aq) + H_2(g)uarr# And we calculate (i) the calcium hydroxide moles... #"Moles of calcium hydroxide"-=(3.70*g)/(74.09*g*mol^-1)=0.0499*mol#... Therefore, (ii) we require an equivalent molar amount of metal. i.e. #0.0499*molxx40.1*g*mol^-1=2.00*g#

Mateo Aguilar · Answer

The mass of calcium is #=2.00g#

The way that calcium and water interact 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#

Olivia Anton · Answer

undefined To calculate the amount of calcium metal needed to produce 3.7g of calcium hydroxide, you need to use stoichiometry. The balanced chemical equation for the reaction between calcium metal ($Ca$) and water ($H_2O$) to produce calcium hydroxide ($Ca(OH)_2$) is:

$$Ca + 2H_2O \rightarrow Ca(OH)_2 + H_2$$

From the equation, you can see that one mole of calcium metal reacts with two moles of water to produce one mole of calcium hydroxide.

1 mole of $Ca(OH)_2$ has a molar mass of approximately $74.09 g/mol$.

Using the molar mass of calcium hydroxide, you can convert the given mass (3.7g) to moles. Then, using the stoichiometry of the reaction, you can determine the moles of calcium metal required. Finally, using the molar mass of calcium, you can convert the moles of calcium to grams.

The calculation steps are as follows:

1. Calculate moles of $Ca(OH)_2$ using its molar mass.
2. Use stoichiometry to find moles of $Ca$ required.
3. Convert moles of $Ca$ to grams.

$$Moles \ of \ Ca(OH)_2 = \frac{3.7g}{74.09g/mol} = 0.05 \ moles $$

$$Moles \ of \ Ca = \frac{0.05 \ moles \ Ca(OH)_2}{1} = 0.05 \ moles $$

$$Grams \ of \ Ca = 0.05 \ moles \times \frac{40.08g}{1 \ mole} = 2.004 \ g $$

So, approximately $2.004 \ g$ of calcium metal must be added to produce $3.7g$ of calcium hydroxide.