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Minchanka [31]
3 years ago
14

An aqueous solution has a mass of 490 grams containing 8.5 × 10-3 gram of calcium ions. The concentration of calcium ions in thi

s solution is
1. 4.3 ppm
2. 8.5 ppm
3. 17 ppm
4. 34 ppm
Chemistry
1 answer:
sammy [17]3 years ago
5 0
To solve the concentration, it is equal to the mass of the calcium ions divided by the total mass of the solution.
C = mass of ions / mass solution
C = ( 8.5 x 10 ^-3 g ) / 490 g
C = 1.7 x 10^-5
since the choices are in ppm, multiply it by 10^6
C = (1.7 x 10^-5 ) 10^6
C = 17 ppm
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CaO(s) + 2 NH4Cl(s) = 2 NH3(g) + H2O(g) + CaCl2(s)
sertanlavr [38]

This is stoichiometry question which involves unit conversions.

Given values:

CaO mass = 11.2g

NH4Cl mass = 22.4g

First convert the two given values to moles.

(11.2g CaO)(1mol CaO / 56.08g CaO) = 0.20mol CaO

(22.4g NH4Cl)(1mol NH4Cl / 53.49g NH4Cl) = 0.42mol NH4Cl

CaO reacts with NH4Cl on a 1:2 ratio.  This means that the reactants will react in this ratio: 0.20mol CaO: 0.40mol NH4Cl.  There will be an excess of 0.02mol NH4Cl unreacted because CaO is the limiting reagent (CaO is used up completely in the reaction and excess NH4Cl remains unreacted).

Now given the equation:

CaO(s) + 2 NH4Cl(s) = 2 NH3(g) + H2O(g) + CaCl2(s)

It is seen that every 2mol of NH4Cl reacts to form 2mol of NH3 because all the elements in the composition of NH3 is found in NH4Cl.  This means that NH4Cl and NH3 are on a 1:1 reaction ratio.  Now we use this relationship:

0.40mol NH4Cl reacted : 0.40mol NH3 produced.  Convert NH3 produced to grams to find solution:

(0.40mol NH3)(17.03g NH3 / 1mol NH3) = 6.8g NH3 produced a)


We determined the excess reactant as 0.02mol NH4Cl in a).  Now we convert it to grams:

(0.02mol NH4Cl)(53.49g NH4Cl / 1mol NH4Cl) = 1.07g NH4Cl unreacted b)



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3 years ago
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Answer:

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Explanation:

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Explanation:

To balance the reactions given, we must understand that the principle to follow is the law of conservation of matter.

Based on this premise, the number of moles of species on the reactant and product side must be the same;

 

      Li   +   Br₂  →    LiBr

Put a,b and c as the coefficient of each species

   aLi  + bBr₂ → cLiBr

balancing Li;

       a  = c

 balancing Br;

        2b  = c

 let a  = 1;

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or a = 2, b = 1 , c = 2

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Using the same method;

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balancing P;

          a  = c

balancing Cl;

          2b  = 3c

let a = 1;

     c  = 1

      b  = \frac{3}{2}

 or

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iii,

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use coefficients a,b,c and d;

    aH₂   +   bSO₂  →   cH₂S   +   dH₂O

balancing H;

    2a = 2c + 2d

balancing S;

     b  = c

balancing O

    2b  = d

 let b  = 1,

      c = 1

       d  = 2

    a  = 3

    3H₂   +   SO₂  →   H₂S   +   2H₂O

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Answer:

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