Answer:
8.8g of Al are necessaries
Explanation:
Based on the reaction, 2 moles of Al are required to produce 3 moles of hydrogen gas.
To solve this question we must find the moles of H2 in 11L at STP using PV = nRT. With these moles we can find the moles of Al required and its mass as follows:
Moles H2:
PV = nRT; PV/RT = n
Where P is pressure = 1atm at STP; V is volume = 11L; R is gas constant = 0.082atmL/molK and T is absolute temperature = 273.15K at STP
Replacing:
1atm*11L/0.082atmL/molK*273.15K = n
n = 0.491 moles of H2 must be produced
Moles Al:
0.491 moles of H2 * (2mol Al / 3mol H2) = 0.327moles of Al are required
Mass Al -Molar mass: 26.98g/mol-:
0.327moles of Al * (26.98g / mol) = 8.8g of Al are necessaries
