Question

If 62.8 g Ca react with 92.3 g HCl according to the reaction below, how many grams of hydrogen gas will be produced, and how many grams of the excess reactant will be left over?Unbalanced equation: Ca + HCl → CaCl2 + H2

Answer 1

2.55 g of hydrogen

12.17 g calcium.

to nearest hundredth.

Step-by-step explanation:

The balanced equation is:

Ca + 2HCl ---> CaCl2 + H2

Using the atomic masses

40.078 g Ca react with 72.916 g of HCl to give 2.016 g HCl

The ratio of Ca to HCl in the above is 1 to 1.81935

so 62.8 g Ca reacts with 62.8 * 1.81935 = 114.245 g HCl

so there is excess of Ca in the given weights.

Therefore the mass of Hydrogen produced

= (2.016 / 72.916) * 92.3

= 2.552 g of hydrogen gas.

The mass of calcium required to produce 2.552 g of hydrogen is:

(2.552 / 2.016) * 40.078

= 50.73 g

So the excess of calcium is 62.8 - 50.73

= 12.17 g.

Answer 2

2.53 grams of hydrogen gas will be produced and 12.2 many grams of the excess reactant i.e. calcium will be left over.

Step-by-step explanation:

`Ca + 2HCl\rightarrow CaCl_2 + H_2`

Moles of calcium =
`(62.8 g)/(40 g/mol)=1.57 mol`

Moles of HCl =
`(92.3 g)/(36.5 g/mol)=2.53 mol`

According to reaction, 2 moles of HCl reacts with 1 mole of calcium :

Then 2.53 moles of HCl will recat with :

`(1)/(2)* 2.53 mol= 1.265 mol` of calcium.

As we can see moles of calcium are in excessive amount. Hence calcium is an excessive reagent.

Moles of calcium left unreacted =1.57 mol - 1.265 mol =0.305 mol

Mass calcium left unreacted = 0.305 mol × 40 g/mol =12.2 g

Since, calcium is an excessive reagent HCl is limiting reagent and the amount of hydrogen gas produced will depend on HCl .

According to reaction, 2 moles of HCl gives 1 mole of hydrogen gas.

Then 2.53 moles of HCl will give:

`(1)/(2)* 2.53 mol= 1.265 mol` of hydrogen gas.

Mass of 1.265 mol of hydrogen gas = 1.265 mol × 2 g/mol = 2.53 g

2.53 grams of hydrogen gas will be produced and 12.2 many grams of the excess reactant i.e. calcium will be left over.