Verified answer

First find the volume of H2 at Standard Temperature and Pressure(STP)

Vi = 1L V2 = ?

P1 = 740 P2 = 760 standard

T1 = 15 + 273 = 288 Kelvin T2 = 273 standard

The equation is P1 x V1 divided by T1 = P2 x V2 divided by T2

Solve for V2 = T2 x P1 xV1 divided by (P2 x T1) Put in numbers

= 273 x 740 x 1 / (760 x 288) = .923 L of H2 produced

Liters of a gas at STP divided by 22.4 = moles of the gas

.923 / 22.4 = .0412 moles of H2

Your equation says that for every 3 moles of H2 produced

2 moles of aluminum is used

So .0412 mole of H2 divided by3 and multiplied by 2 is .0275 moles of Al

Moles x atomic mass = mass in grams

.0275 x 27 = .742 g of Al

First adjust the volume of hydrogen to Standard temp (273 degree K) and pressure (760 mm)

1.0 L x 740/760 mm Hg x 273/288 degree K

divide this answer 0.923 by the molecular volume of a gas (22.4) and you end up with 0.041 moles of H2 gas. It takes 2 moles of Al to make 3 moles of H2 so multiply 0.041 moles H2 by 2/3 to get the number of moles Aluminum. (0.027 moles)

Multiply this by molecular weight of Al (27) and I get 0.729 gm Al

easy man

if u assumed the conversion factor is =1 or 100%

so

1L of H2 by applying the ideal gas law u find #moles of h2

P*V/R*T= n of moles H2 =.0411 mole : P=.973 atm T =288K R=.08206

#moles of h2 *2/3 = # moles of Al needed or prouced =.0275

mass of Al = # moles of Al need or prouced * MW of Al= .0275* MW of Al