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Andrei [34K]
3 years ago
9

What is the empirical formula of a compound containing 5.03 g carbon, 0.42 g hydrogen, and 44.5 g of chlorine

Chemistry
1 answer:
mariarad [96]3 years ago
5 0

Answer:

CHCl₃

Explanation:

We have the following data:

C = 5.03 g

H = 0.42 g

Cl= 44.5 g

First, we divide each mass by the molar mass (MM) of the chemical element to calculate the moles:

MM(C) = 12 g/mol

moles of C = mass/MM(C) = 5.03 g/(12 g/mol) = 0.42 mol C

MM(H) = 1 g/mol

moles of H = mass/MM(H) = 0.42 g/(1 g/mol) = 0.42 mol H

MM(Cl) = 35.4 g/mol

moles of Cl = mass/MM(Cl) = 44.5 g/(35.4 g/mol) = 1.26 mol Cl

Now, we divide the moles by the smallest number of moles (0.42):

0.42 mol C/0.42 = 1 C

0.42 mol H/0.42 = 1 H

1.26 mol Cl/0.42 = 3 Cl

Thus, the C:H:Cl ratio is 1:1:3.

Therefore, the empirical formula is CHCl₃

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5 0
3 years ago
1) When 2.38g of magnesium is added to 25.0cm of 2.27 M hydrochloric acid, hydrogen gas is released.
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Answer:

a. HCl.

b. 0.057 g.

c. 1.69 g.

d. 77 %.

Explanation:

Hello!

In this case, since the reaction between magnesium and hydrochloric acid is:

Mg+2HCl\rightarrow MgCl_2+H_2

Whereas there is 1:2 mole ratio between them.

a) Here, we can identify the limiting reactant as that yielded the fewest moles of hydrogen gas product via the 1:1 and 2:1 mole ratios:

n_{H_2}^{by\  HCl}=0.025L*2.27\frac{molHCl}{1L}*\frac{1molH_2}{2molHCl}  =0.0284molH_2\\\\n_{H_2}^{by\  Mg}=2.38gMg*\frac{1molMg}{24.3gMg}*\frac{1molH_2}{1molMg}=0.0979molH_2

Thus, since hydrochloric yields fewer moles of hydrogen than magnesium, we realize it is the limiting reactant.

b) Here, we use the molar mass of gaseous hydrogen (2.02 g/mol) to compute the mass:

m_{H_2}=0.0284molH_2*\frac{2.02gH_2}{1molH_2}=0.057gH_2

c) Here, we compute the mass of magnesium associated with the yielded 0.0248 moles of hydrogen:

m_{Mg}^{reacted}=0.0284molH_2*\frac{1molMg}{1molH_2}*\frac{24.3gMg}{1molMg}  =0.690gMg

Thus, the mass of excess magnesium turns out:

m_{Mg}^{excess}=2.38g-0.690g=1.69gMg

d) Finally, we compute the percent yield, considering 0.044 g is the actual yield and 0.057 g the theoretical yield:

Y=\frac{0.044g}{0.057g} *100\%\\\\Y=77\%

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