<em>Que quieres clasificar?.... Yo te puedo ayudar si quieres pero dime que quieres clasificar.</em>
Answer:
14.2L at STP
Explanation:
Based on the problem, 2 moles of NH3 produce 6 moles of HF. To solve this question we have to convert the mass of NH3 to moles. With the chemical equation find the moles of HF and using PV = nRT find the liters of HF:
<em>Moles NH3 -Molar mass: 17.031g/mol-</em>
3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3
<em>Moles HF:</em>
0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF
<em>Volume HF</em>
PV = nRT; V = nRT/P
<em>Where V is volume in liters, n are moles of the gas = 0.634 moles, R is gas constant = 0.082atmL/molK, T is absolute temperature = 273.15K at STP and P is pressure = 1atm at STP.</em>
Replacing:
V = 0.634moles*0.082atmL/molK*273.15K / 1atm
V = 14.2L at STP
P1/V1=P2/V2
1/22.4=4/x
X=4 multiple by 22.4
V2=89.6L
