if 35.4 g of Al are treated with 721 mL of 5.86 M HCl, how many gof H gas will t

Question

if 35.4 g of Al are treated with 721 mL of 5.86 M HCl, how many gof H gas will theoretically be formed?
I think I am suppose to be balance the equation but I dontknow what my product would be hence I dont know how to balanceit.
So far I have converted the moles of HCl:
721 mL x (5.86/1000 mL) = 4.22506 = 4.23 x 10^5 mol HCl <--since the 5.86M is the same as mol/L correct?
now i know the answer is suppose to be 3.97 g of H2 gas, buthow in the world does that work? this is like the first practiceprblm in my ch review and i cant seem to work it out-please help meout by guiding me in the right direction. chem is not my forte so iwould love some explanation as to how i can go about these kind ofprblms.
I think I am suppose to be balance the equation but I dontknow what my product would be hence I dont know how to balanceit.
So far I have converted the moles of HCl:
721 mL x (5.86/1000 mL) = 4.22506 = 4.23 x 10^5 mol HCl <--since the 5.86M is the same as mol/L correct?
now i know the answer is suppose to be 3.97 g of H2 gas, buthow in the world does that work? this is like the first practiceprblm in my ch review and i cant seem to work it out-please help meout by guiding me in the right direction. chem is not my forte so iwould love some explanation as to how i can go about these kind ofprblms.

Explanation / Answer

2Al + 6 HCl ----> 2AlCl 3 + 3 H2 Molar mass of Al is = 27 g Molar mass of HCl = 1 + 35.5 = 36.5 g No . of moles of HCl , n = Molarity * Volume                                      = 5.86M * 0.721 L                                      = 4.225 moles No . of moles = mass / Molar mass            4.225 = mass / 36.5             mass = 154.2125 g 2 * 27 g of Al reacts with 6 * 36.5 g of HCl (i.e., ) 54g of Al reacts with 219 g of HCl 35.4 g of Al reacts with X g of HCl X = ( 219 * 35.4 ) / 54     = 143.567 g So, 154.2125 - 143.567 g of HCl left unreacted 54 g of Al produces 3 * 2 g of H2 gas 35.4 g of Al produces Y g of H2 Y = ( 3*2 * 35.4 ) / 54     = 3.9333 g of H2     = Molar mass of Al is = 27 g Molar mass of HCl = 1 + 35.5 = 36.5 g No . of moles of HCl , n = Molarity * Volume                                      = 5.86M * 0.721 L                                      = 4.225 moles No . of moles = mass / Molar mass            4.225 = mass / 36.5             mass = 154.2125 g 2 * 27 g of Al reacts with 6 * 36.5 g of HCl (i.e., ) 54g of Al reacts with 219 g of HCl 35.4 g of Al reacts with X g of HCl X = ( 219 * 35.4 ) / 54     = 143.567 g So, 154.2125 - 143.567 g of HCl left unreacted 54 g of Al produces 3 * 2 g of H2 gas 35.4 g of Al produces Y g of H2 Y = ( 3*2 * 35.4 ) / 54     = 3.9333 g of H2     =

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