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DochEvi [55]
3 years ago
5

Given 3.4 grams of x compound with a molar mass of 85 g and 4.2 grams of y compound with a molar mass of 48 g How much of compou

nd xy can be generated 2x + y2 = 2xy
Chemistry
1 answer:
Ulleksa [173]3 years ago
6 0

Answer:

4.36~g~XY

Explanation:

In this case, we can start with the reaction:

2X + Y_2~->~2XY

If we check the reaction, we will have 2 X and Y atoms on both sides. So, <u>the reaction is balanced</u>. Now, the problem give to us two amounts of reagents. Therefore, we have to find the <u>limiting reagent</u>. The first step then is to find the moles of each compound using the <u>molar mass</u>:

3.4~g~X\frac{1~mol~X}{85~g~X}=0.04~mol~X

4.2~g~Y_2\frac{1~mol~Y_2}{48~g~Y_2}=0.0875~mol~Y_2

Now, we can <u>divide by the coefficient</u> of each compound (given by the balanced reaction):

\frac{0.04~mol~X}{1}=~0.04

\frac{0.0875~mol~Y_2}{2}=0.04375

The smallest value is for "X", therefore this is our <u>limiting reagent</u>. Now, if we use the <u>molar ratio</u> between "X" and "XY" we can calculate the moles of XY, so:

0.04~mol~X\frac{2~mol~XY}{2~mol~X}=0.04~mol~XY

Finally, with the molar mass of "XY" we can calculate the grams. Now, we know that 1 mol X = 85 g X and 1 mol Y_2 = 48 g Y_2 (therefore 1 mol Y = 24 g Y). With this in mind the <u>molar mass of XY</u> would be 85+24 = 109 g/mol. With this in mind:

0.04~mol~XY\frac{109~g~XY}{1~mol~XY}=4.36~g~XY

I hope it helps!

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What are the respective concentrations (M) of Cu2+ and Cl- afforded by dissolving 2.0 g of CuCl2 in water and diluting to 500 mL
Rom4ik [11]

Answer:

M_{Cu^{2+}}=0.030M \\\\M_{Cl^-}=0.060M

Explanation:

Hello.

In this case, the first step is to compute the moles of copper (II) chloride (molar mass: 134.45 g/mol) in 2.0 g as follows:

n_{ClCl_2}=2.0CuCl_2*\frac{1molCuCl_2}{134.45gCuCl_2} =0.015molCuCl_2

Thus, since one mole of copper (II) chloride contains 1 mole of copper (its subscript) and 2 moles of chloride (its subscript), those moles are respectively:

n_{Cu^{2+}}=0.015molCuCl_2*\frac{1molCuCl_2}{1molCuCl_2} =0.015molCu^{2+}\\\\n_{Cl^-}=0.015molCuCl_2*\frac{2molCl^-}{1molCuCl_2} =0.030molCl^-

Therefore, the concentrations (in molar units) considering the volume in liters of the solution (0.500 L for 500 mL) are:

M_{Cu^{2+}}=\frac{0.015molCu^{2+}}{0.500L}=0.030M \\\\M_{Cl^-}=\frac{0.030molCu^{2+}}{0.500L}=0.060M

Best regards.

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3 years ago
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