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devlian [24]
2 years ago
12

If 30g of HCl is dissolved and 10g of NaCl is produced, what is the theoretical yield of the experiment?

Chemistry
1 answer:
givi [52]2 years ago
5 0
HCl + NaOH = NaCl + H₂O

m(HCl)=30g
m'(NaCl)=10g
M(HCl)=36.46 g/mol
M(NaCl)=58.44 g/mol

m(NaCl)/M(NaCl)=m(HCl)/M(HCl)

theoretical mass of sodium chloride
m(NaCl)=M(NaCl)m(HCl)/M(HCl)
m(NaCl)=58.44*30/36.46=48.09 g

the practical yield of the reaction
w=m'(NaCl)/m(NaCl)=m'(NaCl)M(HCl)/M(NaCl)m(HCl)

w=10*36.46/(58.44*30)=0.2080 (20.80%)
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3 years ago
What expression approximates the volume of O2 consumed, measure at STP, when 55 g of Al reacts completely with excess O2?2 Al(s)
nikdorinn [45]

Answer : The volume of O_2 consumed are, 0.75\times 2\times 22.4 L.

Explanation :

The balanced chemical reaction will be:

4Al(s)+3O_2(g)\rightarrow 2Al_2O_3(s)

First we have to calculate the moles of Al.

\text{Moles of }Al=\frac{\text{Mass of }Al}{\text{Molar mass of }Al}

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\text{Moles of }Al=\frac{55g}{27g/mol}=2mole

Now we have to calculate the moles of O_2.

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As, 4 mole of Al react with 3 moles of O_2

So, 2 mole of Al react with \frac{3}{4}\times 2=0.75\times 2 moles of O_2

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3 0
3 years ago
A 1.00 g sample of a metal X (that is known to form X ions in solution) was added to 127.9 mL of 0.5000 M sulfuric acid. After a
Semenov [28]

<u>Answer:</u> The metal having molar mass equal to 26.95 g/mol is Aluminium

<u>Explanation:</u>

  • To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

Molarity of NaOH solution = 0.5000 M

Volume of solution = 0.03340 L

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of NaOH}}{0.03340L}\\\\\text{Moles of NaOH}=(0.5000mol/L\times 0.03340L)=0.01670mol

  • The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 0.01670 moles of NaOH will react with = \frac{1}{2}\times 0.01670=0.00835mol of sulfuric acid

Excess moles of sulfuric acid = 0.00835 moles

  • Calculating the moles of sulfuric acid by using equation 1, we get:

Molarity of sulfuric acid solution = 0.5000 M

Volume of solution = 127.9 mL = 0.1279 L    (Conversion factor:  1 L = 1000 mL)

Putting values in equation 1, we get:

0.5000M=\frac{\text{Moles of }H_2SO_4}{0.1279L}\\\\\text{Moles of }H_2SO_4=(0.5000mol/L\times 0.1279L)=0.06395mol

Number of moles of sulfuric acid reacted = 0.06395 - 0.00835 = 0.0556 moles

  • The chemical equation for the reaction of metal (forming M^{3+} ion) and sulfuric acid follows:

2X+3H_2SO_4\rightarrow X_2(SO_4)_3+3H_2

By Stoichiometry of the reaction:

3 moles of sulfuric acid reacts with 2 moles of metal

So, 0.0556 moles of sulfuric acid will react with = \frac{2}{3}\times 0.0556=0.0371mol of metal

  • To calculate the molar mass of metal for given number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Mass of metal = 1.00 g

Moles of metal = 0.0371 moles

Putting values in above equation, we get:

0.0371mol=\frac{1.00g}{\text{Molar mass of metal}}\\\\\text{Molar mass of metal}=\frac{1.00g}{0.0371mol}=26.95g/mol

Hence, the metal having molar mass equal to 26.95 g/mol is Aluminium

6 0
3 years ago
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