Answer
0.41g Zn
Work Step by Step
Zn + H$_2$SO$_4$ --> ZnSO$_4$ + H$_2$
Given information:
159 mL H$_2$ --> 0.159L
24 degrees Celcius --> 297 K
738 torr --> 0.97 atm
We are asked to calculate the amount of Zn consumed when H$_2$ at the given condition is extracted
Note that from the formula, Zn and H$_2$ have a 1:1 ratio
Using PV = nRT,
n = PV/RT
n = (0.97 x 0.159)/((0.08201 x 297)
n = 0.0063 mol of Zn consumed
Converting this into grams,
0.0063 mol x 65.38 g / 1 mol Zn = 0.41 g Zn
