Question

1.Calculate the mass of HCl required to form 5 moles of CaCl2​

Answer 1

Answer

Number of moles of HCl=250 mL×

1000

0.76 M

​

=0.19 mol

Mass of CaCO

3

​

=1000 g

Number of moles of CaCO

3

​

=

100 g

1000 g

​

=10 mol

According to given equation 1 mol of CaCO

3

​

(s) requires 2 mol of HCl(aq). Hence, for the reaction of 10 mol of CaCO

3

​

(s) number of moles of HCl required would be :

10 mol CaCO

3

​

×

1 mol CaCO

3

​

(s)

2 mol HCl(aq)

​

=20 mol HCl(aq)

But we have only 0.19 mol HCl(aq), hence HCl(aq) is limiting reagent. So, amount of CaCl

2

​

 formed will depend on the amount of HCl available. Since, 2 mol HCl(aq) forms 1 mol of CaCl

2

​

, therefore, 0.19 mol of HCl(aq) would give:

0.19 mol HCl(aq)×

2 mol HCl(aq)

1 mol CaCl

2

​

(aq)

​

=0.095 mol of CaCl

2

​

or 0.095× molar mass of CaCl

2

​

=0.095×111=10.54 g

Explanation:MARK ME THE BRAINLIEST