Respuesta :
Answer:
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.
Explanation:
[tex]Ca + 2HCl\rightarrow CaCl_2 + H_2[/tex]
Moles of calcium = [tex]\frac{62.8 g}{40 g/mol}=1.57 mol[/tex]
Moles of HCl = [tex]\frac{92.3 g}{36.5 g/mol}=2.53 mol[/tex]
According to reaction, 2 moles of HCl reacts with 1 mole of calcium :
Then 2.53 moles of HCl will recat with :
[tex]\frac{1}{2}\times 2.53 mol= 1.265 mol[/tex] 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:
[tex]\frac{1}{2}\times 2.53 mol= 1.265 mol[/tex] 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.
Answer:
2.55 g of hydrogen
12.17 g calcium.
to nearest hundredth.
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.
