How many grams of calcium chloride are needed to produce 1.50 g of potassium chloride? cacl2(aq) + k2co3(aq) → 2 kcl(aq) + caco3(aq) 1.12 g

Question

How many grams of calcium chloride are needed to produce 1.50 g of potassium chloride? cacl2(aq) + k2co3(aq) → 2 kcl(aq) + caco3(aq) 1.12 g 2.23 g 0.896 g 4.47 g?

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2022-04-12T07:33:06+00:00 2 Answers 0

Answers ( 2 )

  1. Delilah
    0
    2022-04-12T07:34:06+00:00

    The mass calcium chloride needed is 1.12 g

    The balanced equation for the reaction is illustrated below:

    CaClu2082 + Ku2082COu2083 -> 2KCl + CaCOu2083

    Next, we shall determine the mass of CaClu2082 that reacted and the mass of KCl produced from the balanced equation.

    Molar mass of CaClu2082 = 40 + (35.5×2)

    = 40 + 71

    = 111 g/mol

    Mass of CaClu2082 from the balanced equation = 1 x 111 = 111 g

    Molar mass of KCl = 39 + 35.5

    = 74.5 g/mol

    Mass of KCl from the balanced equation = 2 x 74.5 = 149 g

    Thus,

    From the balanced equation above,

    111 g of CaClu2082 reacted to produce 149 g of KCl.

    Finally, we shall determine the mass of CaClu2082 needed to produce 1.5 g of KCl as follow:

    From the balanced equation above,

    111 g of CaClu2082 reacted to produce 149 g of KCl.

    Therefore, X g of CaClu2082 will react to produce 1.5 g of KCl i.e

    X g of CaClu2082 = [tex]frac{1.5 * 111}{149}[/tex]

    X g of CaClu2082 = 1.12 g

    Thus, 1.12 g of calcium chloride, CaClu2082 is needed to produce 1.5 g of KCl.

    Learn more: https://brainly.com/question/13143779

  2. Josie97
    0
    2022-04-12T07:34:40+00:00

    The calcium chloride needed to produce 1.5 grams of KCl is: 1.12 grams

    Further explanation

    Stoichiometry in Chemistry learn about chemicals mainly emphasizes quantitative, such as the calculation of volume, mass, number, which is related to numbers, molecules, elements, etc.

    A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

    In the reaction there are also manifestations of reagent substances namely gas (g), liquid (liquid / l), solid (solid / s) and solution (aqueous / aq).

    Reactions that occur:

    CaClu2082 (aq) + Ku2082COu2083 (aq) u2192 2 KCl (aq) + CaCOu2083 (aq)

    From the reaction equation above, the reaction coefficient shows the mole ratio of reagents and products

    mol CaClu2082: moles of KCl = 1: 2

    relative molecular mass KCl = 74.6

    relative molecular mass of CaCl2 = 111

    then the mole:

    mole KCl = mass: relative molecular mass

    mole KCl = 1.5 gram: 74.6

    mole KCl = 0.020107

    so that

    mole CaClu2082 = 1/2. 0.020107

    mole CaClu2082 = 0.0100535

    So the mass of CaCl2 = mole x relative molecular mass

    mass CaClu2082 = 0.0100535 x 111

    mass CaClu2082 = 1.12 gram

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    Keywords: CaCl2, KCl, mole, mass, relative molecular mass

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