Respuesta :
Answer:
0.00203 moles of calcium chloride can be produced.
Explanation:
Molarity of HCl = 0.62 M
Volume of the solution = 6.55 mL = 0.00655 L
Moles of HCl = n
[tex] Molarity=\frac{moles}{Volume (L)}[/tex]
[tex]0.62 M=\frac{n}{0.00655 L}[/tex]
n = 0.004061 mol
Moles of calcium carbonate = [tex]\frac{7.54 g}{100 g/mol}=0.0754 mol[/tex]
[tex]CaCO_3+2HCl\rightarrow CaCl_2+H_2CO_3[/tex]
According to reaction, 1 mol of calcium carbonate reacts with 2 moles of HCl.
Then 0.0754 moles of calcium carbonate will react with:
[tex]\frac{2}{1}\times 0.0754 mol=0.1508 mol[/tex] of HCl
As we can see that number moles required by 0.0754 moles of calcium carbonate is 0.1508 moles of HCl but we only have 0.004061 mol of HCl. This means that HCl is in limiting amount.
So, moles of calcium chloride will depend upon moles of HCl.
According to reaction, 2 mole sof HCl gives 1 mole of calcium chloride.
Then 0.004061 mol of HCl will give:
[tex]\frac{1}{2}\times 0.004061 mol=0.0020305 mol\approx 0.00203 mol[/tex] of calcium chloride
0.00203 moles of calcium chloride can be produced.
