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Wittaler [7]
3 years ago
15

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

ll the metal had reacted, the remaining acid required 0.03340 L of 0.5000 M NaOH solution for complete neutralization. Calculate the molar mass of the metal and identify the element.
Chemistry
1 answer:
Semenov [28]3 years ago
6 0

<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

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Consider the reaction of CaCN2 and water to produce CaCO3 and NH3 according to the reaction CaCN2 + 3H2O → CaCO3 + 2 NH3 . How m
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Answer:

56 g. Option 3.

Explanation:

The reaction is: CaCN₂ + 3H₂O → CaCO₃ + 2 NH₃

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3 moles of water need 1 mol of cianide to react

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8 0
3 years ago
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Answer:

n = 0.573mol

Explanation:

PV = nRT => n = PV/RT

P = 1.5atm

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Read 2 more answers
In a 0.730 M solution, a weak acid is 12.5% dissociated. Calculate Ka of the acid.
Mamont248 [21]

Answer:

Approximately 1.30 \times 10^{-2}, assuming that this acid is monoprotic.

Explanation:

Assume that this acid is monoprotic. Let \rm HA denote this acid.

\rm HA \rightleftharpoons H^{+} + A^{-}.

Initial concentration of \rm HA without any dissociation:

[{\rm HA}] = 0.730\; \rm mol \cdot L^{-1}.

After 12.5\% of that was dissociated, the concentration of both \rm H^{+} and \rm A^{-} (conjugate base of this acid) would become:

12.5\% \times 0.730\; \rm mol \cdot L^{-1} = 0.09125\; \rm mol \cdot L^{-1}.

Concentration of \rm HA in the solution after dissociation:

(1 - 12.5\%) \times 0.730\; \rm mol \cdot L^{-1} = 0.63875\; \rm mol\cdot L^{-1}.

Let [{\rm HA}], [{\rm H}^{+}], and [{\rm A}^{-}] denote the concentration (in \rm mol \cdot L^{-1} or \rm M) of the corresponding species at equilibrium. Calculate the acid dissociation constant K_{\rm a} for \rm HA, under the assumption that this acid is monoprotic:

\begin{aligned}K_{\rm a} &= \frac{[{\rm H}^{+}] \cdot [{\rm A}^{-}]}{[{\rm HA}]} \\ &= \frac{(0.09125\; \rm mol \cdot L^{-1}) \times (0.09125\; \rm mol \cdot L^{-1})}{0.63875\; \rm mol \cdot L^{-1}}\\[0.5em]&\approx 1.30 \times 10^{-2} \end{aligned}.

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