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What volume of hydrogen at STP is produced from the reaction of 50.0 g of Mg and the equivalent of 75 g of HCl?
Wiki User
∙ 12y ago
Zn + 2HCl ---> H2 + ZnCl2
n = 0.112
mw = 65.39
g = 7.35
From these values (they are for the zinc), you can see that only .112 mol of Zn reacted.
To find the number of moles of HCl, we use n = cV.
n = 1.2 x 0.5
= 0.6 x 2 mols
= 1.2 mol HCl.
From this we can see that zinc is the limiting reagent and that 0.112 mol of H2 gas was produced.
g = mw x n
g = 2.016 x 0.112
= 0.225792 grams.
Wiki User
∙ 14y ago
Wiki User
∙ 13y agoBalanced equation first.
Zn + 2HNO3 >> Zn(NO3)2 + H2
( I forgot STP conversion, so I will use the trusty PV = nRT )
(1atm)(75.5l) = n(0.08206Latm/molK)(273.15K)
n (moles) = 3.37 moles H2
3.37 moles H2 (1mol Zn/1mol H2)(65.41g Zn/1mol Zn)
= 220.47 grams Zn needed.
Wiki User
∙ 13y agoBalanced equation.
2Al + 6HCl -> 2AlCl3 + 3H2
find moles hydrogen gas
2.70 grams Al (1 mole Al/26.98 grams)(3 moles H2/2 mole H2)
= 0.10007 moles H2
Now use PV = nRT
(1 atm)(volume H2) = (0.10007 moles H2)(0.08206 L*atm/mol*K)(298.15 K)
= 2.45 Liters of hydrogen gas
------------------------------------
Wiki User
∙ 12y ago2.50 g Zn * (1 mol Zn/65.39 g Zn) * (1 mol H2/ 1 mol Zn) * ( 22.4 L H2/1 mol H2) = 0.856 L of H2
Wiki User
∙ 12y agoThe answer is 23L. There is 25.0 g of excess reagent that is left over in that problem.
Wiki User
∙ 12y ago23
Wiki User
∙ 10y ago14.9cm^3
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