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Anonymous asked in Science & MathematicsChemistry · 7 years ago

When 350 mL of a 0.6 M solution of calcium nitrate is combined with 400 mL of a 1.4 M solution of sodium ca?

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7. When 350 mL of a 0.6 M solution of calcium nitrate is combined with 400 mL of a 1.4 M solution of sodium carbonate, a precipitate forms. How many grams of precipitate are obtained?

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  • 7 years ago
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    Ca(NO3)2 + Na2CO3 --> CaCO3 + 2NaNO3

    1. Find out the amount of moles of each starting reactant

    0.6moles/L *(0.350L) = 0.21 moles Ca(NO3)2

    1.4moles/L * (0.400L) = 0.56 moles Na2CO3

    2. Find out which one is the limiting reagent (which gives you a lesser amount of precipitate)

    0.21 moles Ca(NO3)2 * (1 mole CaCO3/ 1 mole Ca(NO3)2) * (100.086 g CaCO3/1 mole CaCO3) = 21.1 grams

    0.56 moles Na2CO3* (1 mole CaCO3/ 1 mole Na2CO3) * (100.086 g CaCO3/1 mole CaCO3) = 56.0 grams

    The answer should be in 2 S.F. so 21 grams of CaCO3

    Source(s): Chemistry tutor
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