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9 June, 14:02

How many milliliters of 0.260 m na2s are needed to react with 35.00 ml of 0.315 m agno3?

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  1. 9 June, 14:28
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    The complete balanced chemical reaction is:

    2 AgNO3 + Na2S - - > 2 NaNO3 + Ag2S

    First let us calculate the number of moles of AgNO3.

    moles AgNO3 = 0.315 M * 0.035 L

    moles AgNO3 = 0.011025 mol

    From the reaction, 1 mole of Na2S is needed for every 2 moles of AgNO3 hence:

    moles Na2S required = 0.011025 mol AgNO3 * (1 mol Na2S / 2 mol AgNO3)

    moles Na2S required = 5.5125 x 10^-3 mol

    Therefore volume required is:

    volume Na2S = 5.5125 x 10^-3 mol / 0.260 M

    volume Na2S = 0.0212 L = 21.2 mL
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