How man Liters are there in 450 grams of NH3 gas at STP?

When using mass values from the periodic table round to 2 decimal places. EX: H-1.008 g > 1.01 g

There are 591,18 liters of NH3 gas at STP.

Explanation:

We use the ideal gas formula, with the constant R = 0.082 l atm / K mol. First, we calculate the number of moles of NH3 (we obtain the mass of 1 mol from the atomic weights of the elements of the periodic table).The STP conditions are : 1 atm pressure and 273 K temperature. Solve for the formula, V (volume):

Weight 1 mol NH3= Weight N + (Weight H)x3= 14,01 g + 3 x 1,01 g=17,04 g/mol

17,04 g ---1 mol NH3

450 g--x= (450 g x 1 mol NH3)/17,04g= 26,41 mol NH3

PV=nRT ---> V= (nRT)/P

V=(26,41 mol x 0.082 l atm / K mol x 273 K)/ 1 atm

V=591, 18 L

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as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2

i think it's letter a.