The experiment described in the procedure was performed using a sample of Alumin
ID: 561683 • Letter: T
Question
The experiment described in the procedure was performed using a sample of Aluminun with HCl(aq). The lab conditions were 25.00 °C, and 747.00 torr barometric pressure. 123.5 mg of Aluminum (MM:26.98 g/mol) was reacted with 25.00 mL of 1.00 HCl(aq). 176.68g of water was collected when the reaction was complete, and it required 15.23 mL of 0.750 M NaOH(aq) to tirarte the reaction mixture. Water at 25.00 °C has a density of 0.9970 g/cm^3, and a vapor pressure of 23.80 torr. The value R is 0.082058 L•atm/mol•K. How many moles of HCl(aq) are consumed in the reaction with the metal? The experiment described in the procedure was performed using a sample of Aluminun with HCl(aq). The lab conditions were 25.00 °C, and 747.00 torr barometric pressure. 123.5 mg of Aluminum (MM:26.98 g/mol) was reacted with 25.00 mL of 1.00 HCl(aq). 176.68g of water was collected when the reaction was complete, and it required 15.23 mL of 0.750 M NaOH(aq) to tirarte the reaction mixture. Water at 25.00 °C has a density of 0.9970 g/cm^3, and a vapor pressure of 23.80 torr. The value R is 0.082058 L•atm/mol•K. How many moles of HCl(aq) are consumed in the reaction with the metal? The experiment described in the procedure was performed using a sample of Aluminun with HCl(aq). The lab conditions were 25.00 °C, and 747.00 torr barometric pressure. 123.5 mg of Aluminum (MM:26.98 g/mol) was reacted with 25.00 mL of 1.00 HCl(aq). 176.68g of water was collected when the reaction was complete, and it required 15.23 mL of 0.750 M NaOH(aq) to tirarte the reaction mixture. Water at 25.00 °C has a density of 0.9970 g/cm^3, and a vapor pressure of 23.80 torr. The value R is 0.082058 L•atm/mol•K. How many moles of HCl(aq) are consumed in the reaction with the metal?Explanation / Answer
2Al(s) + 6HCl(AQ) ---> 2AlCl3(s) + 3H2(g)
no of mol of Al reacted = 0.123/27 = 0.00455 mol
no of mol of HCl == 25*1/1000 = 0.025 mol
limiting reactant = Al
NO of mol of naOH reacted =15.23*0.75/1000 = 0.0114 mol
no of mol of HCl consumed = 0.025 - 0.0114 = 0.0136 mol
Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.