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26 December, 01:34

A 1.540 gram sample of an alloy containing only tin and zinc was reacted with excess fluorine gas to produce 2.489 grams in total of a mixture of tin IV fluoride and zinc fluoride. Calculate the percent composition by mass of the two metals in the alloy.

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  1. 26 December, 02:57
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    Tin: 54.3%

    Zinc: 45.7%

    Explanation:

    The molar masses of the elements are:

    Tin: Sn = 117.710 g/mol

    Zinc: Zn = 65.409 g/mol

    Fluorine: F = 18.998 g/mol

    The fluorine gas in excess, so the reaction consumes all the alloy, and all the tin is converted to SnF₄ and all the zinc is converted to ZnF₂. The molar masses of the fluorides are:

    SnF₄ = 117.710 + 4*18.998 = 193.702 g/mol

    ZnF₂ = 65.409 + 2*18.998 = 103.405 g/mol

    If we call x the number of moles of SnF₄, and y the number of moles of ZnF₂, the total mass can be calculated knowing that the mass is the number of moles multiplied by the molar mass:

    193.702x + 103.405y = 2.489

    The number of moles of Sn is the same as SnF₄ (1:1), and also the number of moles of Zn is the same as ZnF₂ (1:1), so the mass of the alloy:

    117.710x + 65.409y = 1.540

    if we multiply it by - 1.581 and sum with the other equation:

    117.710x * (-1.581) + 65.409y * (-1.581) + 193.702x + 103.405y = 1.540 * (-1.581) + 2.489

    7.60249x = 0.05426

    x = 0.0071 mol of Sn

    117.710*0.0071 + 65.409y = 1.540

    65.409y = 0.704259

    y = 0.0108 mol of Zn

    The masses are the molar mass multiplied by the number of moles:

    Sn: 117.710*0.0071 = 0.836 g

    Zn: 65.409*0.0108 = 0.704 g

    The percent composition is the mass of the substance divided by the total mass multiplied by 100%:

    Sn: (0.836/1.540) * 100% = 54.3%

    Zn: (0.704/1.540) * 100% = 45.7%
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