Question

How many grams of ca are needed to react completely with 2.20 L of a 4.50 m hcl solution

Answer 1

Ca + 2HCl = CaCl₂ + H₂

c=4.50 mol/l
v=2.20 l

n(HCl)=cv

m(Ca)/M(Ca)=n(HCl)/2

m(Ca)=M(Ca)cv/2

m(Ca)=40g/mol·4.50mol/l·2.20l/2=198 g

198 grams of Ca are needed

Answer 2

Answer:

There are needed 198g of Ca to react completely with 2.20L of a 4.50M HCl solution

Explanation:

First we need to establish the chemical reaction between Ca and HCl.

Ca + HCl → CaCl₂ + H₂

Then we need to balance the equation to find out the stoichiometry of the reaccion, to do this we need to keep in count that in a chemical reaction mass cannot be generated or created, so the reactants must have the same amount of mass as the product

The balanced equation is:

Ca + 2HCl → CaCl₂ + H₂

This means that 1 mol of Ca reacts with 2 moles of HCl to form 1 mol of CaCl₂ and 1 mol of H₂.

Next we need to find out how many moles we have in the solution.

1L of solution ⇒ 4.50 moles of HCl

2.20L of solution ⇒ x moles of HCl

$x = \frac{2.20L · 4.50 mol}{1L}$

x = 9.9 moles of HCl

Now we use the stoichiometric ratio between HCl a Ca to find out how many moles react with 9.9 moles of HCl.

2 moles of HCl  ⇒ 1 mol of Ca

9.9 moles of HCl ⇒ x moles of Ca

$x = \frac{9.9 mol· 1 mol}{2mol}$

x = 4.95 moles of Ca

Finally we calculate how many grams 4.95 moles of Ca are, using the Ca molar mass.

1 mol of Ca ⇒ 40g

4.95 moles of Ca ⇒ x g

$x = \frac{4.95 mol ·40g}{1 mol}$

x = 198g