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monitta
3 years ago
10

Magnesium metal reacts with iodine gas at high temperatures to form magnesium iodide. what mass of mgi2 can be produced from the

reaction of 5.15 g mg and 50.0 g i2
Chemistry
1 answer:
sergiy2304 [10]3 years ago
5 0
<span>54.8 g of MgI2 can be produced. To solve this, you need to determine the molar mass of each reactant and the product. First, look up the atomic weights of iodine and magnesium Atomic weight of Iodine = 126.90447 Atomic weight of Magnesium = 24.305 Molar mass of MgI2 = 24.305 + 2 * 126.90447 = 278.11394 Now determine how many moles of Iodine and Magnesium you have moles of Iodine = 50.0 g / 126.90447 g/mol = 0.393997154 mole moles of Magnesium = 5.15 / 24.305 g/mol = 0.211890557 mole Since for every magnesium atom, you need 2 iodine atoms and since the number of moles of available iodine isn't at least 2 times the available moles of magnesium, iodine is the limiting reagent. So figure out how many moles of magnesium will be consumed by the iodine 0.393997154 mole / 2 = 0.196998577 mole. This means that you can make 0.196998577 moles of MgI2. Now simply multiply by the previously calculated molar mass of MgI2 0.196998577 mole * 278.11394 g/mole = 54.78805 g Round the result to the correct number of significant figures. 54.78805 g = 54.8 g</span>
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Explanation:

8 0
2 years ago
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.

8 0
3 years ago
Give the nuclide symbol for an atom that has<br> mass number 64 and 35 neutrons.
azamat
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In ground state, number of protons in an atom is equal to the number of electrons. Therefore,
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Using the periodic table, we will find that the element that has 29 electrons in ground state is copper.
The nuclide symbol of copper is shown in the attached image.

5 0
3 years ago
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Elenna [48]

Answer:

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explanation

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