5F248+ 2NH3(8) ► NF418 + 6HF a) If you start with 3.6 g of NH, and excess F2, how many liters of HF at STP can you make?

Question

5F248+ 2NH3(8) ► NF418 + 6HF
a) If you start with 3.6 g of NH, and excess F2, how many liters of HF at STP can you make?

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Doris 3 years 2021-07-31T19:13:46+00:00 1 Answers 10 views 0

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  1. Vodka
    0
    2021-07-31T19:15:29+00:00
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    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:

    Moles NH3 -Molar mass: 17.031g/mol-

    3.6g NH3 * (1mol / 17.031g) = 0.211 moles NH3

    Moles HF:

    0.211 moles NH3 * (6mol HF / 2mol NH3) = 0.634 moles HF

    Volume HF

    PV = nRT; V = nRT/P

    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.

    Replacing:

    V = 0.634moles*0.082atmL/molK*273.15K / 1atm

    V = 14.2L at STP

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )