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Stels [109]
3 years ago
5

1. Calculate the mass (in grams) of 3.8 moles of the following:

Chemistry
1 answer:
muminat3 years ago
4 0

Explanation:

Given;

  Number of moles of the elements = 3.8moles

Molar mass of Gold  = 196.966g/mol

Molar mass of Magnesium  = 24.3g/mol

Molar mass of Tin  = 118.710g/mol

Molar mass of Bromine  = 79.904g/mol

   Mass  = number of moles x molar mass

Mass of Gold = 3.8 x 196.966  = 748.5g

 Mass of Magnesium  = 3.8 x 24.3  = 92.3g

Mass of Tin  = 3.8 x 118.710  = 451.1g

Mass of Bromine  = 3.8 x 79.904  = 303.6g

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If 28.0 grams of Pb(NO3)2 react with 18.0 grams of NaI, what mass of PbI2 can be produced? Pb(NO3)2 + NaI → PbI2 + NaNO3
ss7ja [257]

Answer:- 27.7 grams of PbI_2 are produced.

Solution:- The balanced equation is:

Pb(NO_3)_2+2NaI\rightarrow PbI_2+2NaNO_3

let's convert the grams of each reactant to moles and calculate the grams of the product and see which one gives least amount of the product. This least amount would be the answer as the least amount we get is from the limiting reactant.

Molar mass of Pb(NO_3)_2 = 207.2+2(14.01)+6(16)  = 331.22 gram per molmolar mass of NaI = 22.99+126.90 = 149.89 gram per molMolar mass of [tex]PbI_2 = 207.2+2(126.90) = 461 gram per mol

let's do the calculations for the grams of the product for the given grams of each of the reactant:

28.0gPb(NO_3)_2(\frac{1molPb(NO_3)_2}{331.22gPb(NO_3)_2})(\frac{1molPbI_2}{1molPb(NO_3)_2})(\frac{461gPbI_2}{1molPbI_2})

= 39.0gPbI_2

18.0gNaI(\frac{1molNaI}{149.89gNaI})(\frac{1molPbI_2}{2molNaI})(\frac{461gPbI_2}{1molPbI_2})

= 27.7gPbI_2

From above calculations, NaI gives least amount of PbI_2, so the answer is, 27.7 g of PbI_2 are produced.

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Answer:

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Additionally, this compound has 2 substituents (chloros) so you have to indicate this number (di).

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See attached figure for the structure.

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