The molar mass of Al(OH)3 = 27+(16+1)*3 = 78 g/mol. Dividing the 200.0 g by the molar mass of 78 g/mol = 2.564 moles. From the balanced equation, 2 moles of Al(OH)3 are equivalent to 1 mole of Al2(SO4)3, so if 2.564 moles of Al(OH3) are used, we divide by 2 to find that 1.282 moles of Al2(SO4)3 are formed. The molar mass of Al2(SO4)3 is 27*2+(32.07+16*4)*3 = 342.21 g/mol, so multiplying this by 1.282 moles gives 438.71 grams of aluminum sulfate produced.
Answer is: 15,21 liters of nitrogen is needed.
Chemical reaction: N₂ + 3H₂ ⇄ 2NH₃.
V(H₂) = 45,0 L.
V(N₂) = ?
n(H₂) = V(H₂) ÷ Vm.
n(H₂) = 45 L ÷ 22,4 L/mol
n(H₂) = 2,01 mol.
from chemical reaction: n(N₂) : n(H₂) = 1 : 3.
n(N₂) = 0,67 mol.
V(N₂) = 0,67 mol · 22,4 L/mol.
V(N₂) = 15,21 L.
Under STP 1 mol of gas has volume 22.4 L.
2C2H2 + 5O2 ------> 4CO2 + 2H2O
from reaction 5*22.4 L 4 *22.4L
given 60L x L
x=(60*4 *22.4)/5*22.4 = (60*4)/5 =48 L of CO2