A) I need 50ml of 2.5 m HCL soliton for my experiment my lab only has 6 m HCL av
ID: 928820 • Letter: A
Question
A) I need 50ml of 2.5 m HCL soliton for my experiment my lab only has 6 m HCL available how do I prepare it?B) balance the equation al + HCL-> alcl3 + h2
C). When 10g of al reacts with excess HCL all of the h2 gas is captured in a 2 L cointainer what will be the pressure at 300k
D) 300 ml of 2.5 naoh is required to neutralize 500 ml what is the concentration
E). Balance the equation h2 + o2 _> h20
What is the limiting reaction if 100 g of h2 are reacted with 50g o2
A) I need 50ml of 2.5 m HCL soliton for my experiment my lab only has 6 m HCL available how do I prepare it?
B) balance the equation al + HCL-> alcl3 + h2
C). When 10g of al reacts with excess HCL all of the h2 gas is captured in a 2 L cointainer what will be the pressure at 300k
D) 300 ml of 2.5 naoh is required to neutralize 500 ml what is the concentration
E). Balance the equation h2 + o2 _> h20
What is the limiting reaction if 100 g of h2 are reacted with 50g o2
B) balance the equation al + HCL-> alcl3 + h2
C). When 10g of al reacts with excess HCL all of the h2 gas is captured in a 2 L cointainer what will be the pressure at 300k
D) 300 ml of 2.5 naoh is required to neutralize 500 ml what is the concentration
E). Balance the equation h2 + o2 _> h20
What is the limiting reaction if 100 g of h2 are reacted with 50g o2
Explanation / Answer
A. Since, only 6M HCl is available in the lab. The only solution is to dilute the 6M solution to obtain the desired dilution. Use the M1V1=M2V2 to solve this problem.
Here M1= 6M, V1=?, M2= 2.5M, V2= 50 mL.
6* V1= 2.5*50, therefore V1= 20.83 mL. Dilute 20.83 mL of 6M HCl to obtain 50mL of 2.5 M HCl.
B. To balance this equation, Al+ HCl ----> AlCl3+ H2, On the RHS, 3 Cl atoms are required and 2 H atoms. Hence,
2Al+ 6HCl ------> 2AlCl3 + 3H2
C. Given is Al + HCl ------> AlCl3 + H2 and a balanced equation for this is 2Al+ 6 HCl-------> 2AlCl3+ 3H2
Here, let us first determine the no of moles of the gas produced.
From the balanced equation 2 moles of aluminium produces 3 moles of hydrogen i.e the mole ratio is 2:3
10g Al* 1mol/ 27g * 3 mol H2/ 2 mol Al= 0.5556 mol H2.
From the Ideal gas equation, P=?, V= 2L, n= 0.5556, R= 0.08216 L atm/mol K, T= 300 K
P= nRT/V= (0.5556*0.08216*300)/ 2=6.8472 atm
D. Since the acid is not given let us consider the acid in question to be monobasic i.e. HX and it dissociates in water completely. So if 1 mole of acid is dissociated, then one mole of H+ is present in the aqueous solution.
300 mL of 2.5 M NaOH dissociates to give x moles of NaOH= 2.5 * 0.3= 0.75 moles
The ratio between NaOH and the monobasic acid is 1:1. Therefore moles of NaOH= moles of HX= 0.75
Hence, concentration of monobasic acid is, Molarity= no of moles/ volume in liters= 0.75/ 0.5= 1.5 M.
E: H2+O2 ------> H2O Here 2 H atoms are balanced but the O is not hence, the balanced reacttion is
2H2+ O2 -----> 2 H2O
It is clear from the equation that twice the number of O atoms are required to form the water molecule. Hence, if one mole of oxygen is present then two moles of hydrogen are required for the formation of water. Thus, hydrogen is used up faster than oxygen and hence, here hydrogen is the limiting factor.