Ap Chem Problem #3?

Question

A 1.000-g sample of a metal chloride, MCl2, is dissolved in water and treated with excess aqueous silver nitrate. The silver chloride that formed weighed 1.286 g. Calculate the atomic mass of M. A) 222.8 g B) 76.00 g C) 152.0 g D) 304.0 g E) none of these

Answer 1

Your answer is (C)

Your reaction is:

MCl2 + AgNO3 (excess) -------> AgCl

According to the problem statement, the metal chloride is the limiting reagent, so you can assume that all of the chlorine in the metal chloride is used up to make the AgCl.

Take the 1.286 g of AgCl (molar mass = 143.321 g/mole) and calculate the number of moles of product you have made:

1.286-g / 143.321-g/mole = 0.009 mole of AgCl as product.

This means that in the product you have 0.009 moles of Ag and 0.009 moles of Cl. Since your only source of Cl is the metal chloride and all of the metal chloride is converted to product, you now know that you started with 0.009 moles of Cl in MCl2.

In order to balance the left hand side of the equation for chlorine, if we have a total of 0.009 mol of Cl, that means we had 0.0045 mol of Cl2. The ratio of metal to Cl2 is 1:1, so you had 0.0045 mole of metal as well. We can use this information to determine the mass of M.

Let m = mass of MCl2, x = # of moles of MCl2, mM = mass of M and mCl = mass of Cl2. We know m, x, and mCl, so we now just need to solve for mM.

You end up with an algebraic equation that look like this because the mass of the molecule is the sum of the molecular masses multiplied by the number of moles of the molecule:

m = x (mM + MCl)

1 g = 0.0045 mole (mM + 70.906 g/mole)

222.22 g/mole = mM + 70.906 g/mole

mM = 151.32 g/mole, so your answer is C

Answer 2

First find the moles of AgCl, which has a relative molar mass Mm = 143.5.

n = m/Mm n = 1.286/143.5, n = 8.96x10^(-3) mol or 0.009 mol (approx).

According to the chem. equation:

MCl2 + 2AgNO3 --> M(NO3)2 + 2AgCl

2 mol of AgCl are produced from 1 mol of MCl2
0.009 mol of AgCl are produced from n = 0.0045 mol of MCl2

Now, you can find the molar mass of MCl2:

m = n x Mm => Mm = m / n, Mm = 1 / 0.0045, Mr = 223.

If Am the atomic mass of metal M, then

Am + 2 x 35.5 = 223 (where 35.5 the atomic mass of chlorine)

Am = 223 - 71, Am = 152

So the right answer is (c): 152 g/mol

Answer 3

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Answer 4

This is not to be used to find answers for homework.

In order to solve your problem you need to find the number of moles of MCl2 used, then work out what weight was M.

To do this, first find the number of moles of Silver Chloride produced.

Answer 5

C