SUP DOG 發問於 科學及數學化學 · 1 十年前

數條chem計數問題 高手幫幫忙! (20 marks)

1. amminia is produced :

N2 (g) + 3H2 (g) --> 2NH3(g)

a. calculate the theoretical yield of ammonia produced from 72.8g of nitrogen

2. a compound X contains 54.5% carbon 9.10% hydrogen and 36.4% oxygen

its relative molecular mass is 88.0 (relative atomic masses: H =1.0 C=12.0 O=16.0)

a. find its empirical formula

b. find its molecular formula

4 個解答

評分
  • 1 十年前
    最愛解答

    1) N2(g) + 3H2(g) --> 2NH3(g)

    From the equation, we can describe in words that:

    1 mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce 2 moles of ammonia gas.

    So suppose there are excess supply of hydrogen gas, i.e. nitrogen gas is the limiting reactant:

    Molar mass of nitrogen (N2) = 2 x 14 = 28g

    So mo. of moles of nitrogen gas in 72.8g = 72.8/28 = 2.6 moles

    So to speak, if the reaction is complete, a maximum of 2.6 x 2 = 5.2 moles of ammonia gas will be produced.

    And this amount is equal to 5.2 x (14 + 3 x 1) = 88.4g

    or 5.2 x 24 = 124.8 dm^3 measured under room temperature and pressure.

    2)

    a) Suppose there is 100g of such compound. Then,

    mass of C = 54.5g = 54.5/12 = 4.54 moles

    mass of H = 9.1g = 9.1/1 = 9.1 moles

    mass of O = 36.4g = 36.4/12 = 2.275 moles

    Thus C : H : O = 4.54 : 9.1 : 2.275 and approximates to

    C : H : O = 2 : 4 : 1

    So the empirical formula is C2H4O

    b)

    Its empirical mass = 2 x 12 + 4 x 1 + 16 = 44

    Hence its molecular mass is twice its empirical mass and then the molecular formula is double of the empirical formula, i.e.

    C4H8O2

    資料來源: My chemical knowledge
  • 1 十年前

    1. In this reaction, we assumed all the nitrogen used can be 100% converted to ammonia. Since we have 72.8 gm of nitrogen molecule, the mole of nitorgen atom is 72.7/ 28 x 2 = 5.2 mol. THis 5.2 mol of ammonia will be completely converted to Ammonia (MW = 17). So, the weight of ammonia formed = 5.2 x 17 = 88.4gm.

    2a. The empirical formula is C= 54.5/12; H = 9.1/1; O=36.4/16 = 4.54: 9.1:2.275 = 2:4:1

    So, the empirical formula should be C2H4O.

    2b. The molecular formula is C: 54.5%x88/12 = 4; H: 9.1%x88/1 = 8 and O: 36.4%x88/16 = 2. So, the molecular formula is C4H8O2.

  • 1 十年前

    1a. N2(g) + 3H2(g) --> 2NH3(g)

    mole ratio of N2 : H2 : NH3 = 1 :3 : 2

    no. of mole of 72.8g of nitrogen =

    mass / molar mass = 72.8/14= 5.2mol

    According to the mole ratio of the formula. N2 : NH3 = 1 : 2

    Therefore the theorectical yeild of NH3 is 5.2 x 2 = 10.4mol

    10.4 mol of NH3: mole x molar mass= mass= 10.4x(14+1X3)= 176.8g

    2a. Epirical formula is

    C= 54.5%

    H= 9.1 %

    O= 36.4%

    Let the compound is 100g

    C= 54.5g; H=9.1g; O=36.4g

    C : H : O

    mol= 54.5g/12; 9.1g/1 ; 36.4g/16

    4.54 : 9.1 :2.2

    mol ratio=4.54/2.2: 9.1/2.2: 36.4/2.2

    2.06 : 4.14 : 1

    therefore the empirical formula is C: H: O = 2:4:1= C2H4O

    b. molecular mass is 88.0

    molecular mass of the empirical formula is (12x2 +1x4 + 16) = 44

    88/ 44 =2

    the molecular formula of the compound X is

    (C2H4O) 2 = C4H8O2

  • 1 十年前

    Sorry,I only know Q2...

    2a. You should assume that there is 100g of compound.That means there are 54.5 g of carbon,9.1g of hydrogen and etc.

    C H O

    Mass: 54.5 9.1 36.4

    Molar mass: 12 1 16

    Number of mole: 4.542 9.1 2.275

    Divided by the smallest number of mole

    Mole ratio: 1.996(~2) 4 1

    So,the empirical formula: C2H4O

    b. Let the molecular formula of the compound be (C2H4O)n,where n is a whole number.

    (C2H4O)n=88

    (12X2+1X4+16)n=88

    44n=88

    n=2

    So,the molecular formula of compound X is C4H8O2.

    Hope I can help u~~~

    資料來源: 本人
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