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cestrela7 [59]
3 years ago
8

Hydrochloric acid reacts with calcium to form hydrogen and calcium chloride. if 100 grams of hydrochloric acid reacts with 100 g

rams of calcium, what is the limiting reactant? 2hcl ca → cacl2 h2 calcium hydrogen calcium chloride hydrochloric acid
Chemistry
2 answers:
HACTEHA [7]3 years ago
8 0

Answer: hydrochloric acid

Explanation: 2HCl+Ca\rightarrow CaCl_2+H_2

As can be seen from the chemical equation, 2 moles of hydrochloric acid react with 1 mole of calcium.

According to mole concept, 1 mole of every substance weighs equal to its molar mass.

Thus 2\times 36.5=73 g of of hydrochloric acid reacts with 40 g of calcium.

100 g of hydrochloric acid react with=\frac{40}{73}\times 100=54.79 gof calcium

Thus HCl is the limiting reagent as it limits the formation of product and calcium is the excess reagent as (100-54.79)=45.21 g is left unreacted.

natita [175]3 years ago
4 0
The balanced chemical reaction is:

2HCl + Ca = CaCl2 + H2

We are given the amount of the reactants to be used for the reaction. These values will be the starting point of our calculations.

100 g HCl ( 1 mol HCl / 36.46 g HCl ) = 2.74 mol HCl

100 g Ca ( 1 mol Ca / 40.08 g ) = 2.08 mol Ca

From the reaction, the mole ratio of the reactants is 2:1 where every 2 moles of hydrochloric acid, 1 mole of calcium is required. Therefore, the limiting reactant for this case is calcium.
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Answer:

1.13moles

Explanation:

Given parameters:

Number of atoms  = 6.777 x 10²³ atoms

Unknown:

Number of moles  = ?

Solution:

A mole of a substance contains the avogadro's number of particles

       6.02  x  10²³ particles   = 1 mole

      6.777 x 10²³ atoms will contain \frac{ 6.777 x 10^{23}  }{6.02 x 10^{23} }   = 1.13moles

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What is the lowest point of a wave called?<br><br> Amplitude<br> Crest<br> Frequency<br> Trough
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Trough

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Aspirin (acetylsalicylic acid, C9H8O4) is a weak monoprotic acid. To determine its acid-dissociation constant, a student dissolv
Triss [41]

Answer:

1. 3.57\times 10^{-4}was the K_a value calculated by the student.

2. 5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

Explanation:

1.

The pH value of Aspirin solution = 2.62

pH=-\log[H^+]

[H^+]=10^{-2.62}=0.00240 M

Moles of s asprin = \frac{2.00 g}{180 g/mol}=0.01111 mol

Volume of the solution = 0.600 L

The initial concentration of Aspirin  = c = \frac{0.01111 mol}{0.600 L}=0.0185 M

HAs\rightleftharpoons As^-+H^+

initially

c       0    0

At equilibrium

(c-x)      x   x

The expression of dissociation constant :

K_a=\frac{[As^-][H^+]}{[HAs]}:

K_a=\frac{x\times x }{(c-x)}

=\frac{0.00240 M\times 0.00240 M}{(0.0185-0.00240 )}

K_a=3.57\times 10^{-4}

3.57\times 10^{-4}was the K_a value calculated by the student.

2.

The pH value of ethylamine = 11.87

pH+pOH=14

pOH=14-11.87=2.13

pOH=-\log[OH^-]

[OH^-]=10^{-2.13}=0.00741 M

The initial concentration of ethylamine = c = 0.100 M

C_2H_5NH_2+H_2O\rightleftharpoons C_2H_5NH_3^{+}+OH^-

initially

c                    0    0

At equilibrium

(c-x)                x   x

The expression of dissociation constant :

K_b=\frac{[C_2H_5NH_3^{+}][OH^-]}{[C_2H_5NH_2]}:

K_b=\frac{x\times x}{(c-x)}

=\frac{0.00741\times 0.00741}{(0.100-0.00741)}

K_b=5.93\times 10^{-4}

5.93\times 10^{-4}was the K_b of ethylamine value calculated by the student.

3 0
3 years ago
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