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3 years ago

Number of grams of hcl that can react with 0.500 grams of Al(OH)3

Chemistry

1 answer:

vlabodo [156]3 years ago

7 0

Answer:

0.7g of HCl

Explanation:

First, let us write a balanced equation for the reaction between HCl and Al(OH)3.

This is illustrated below:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

Next, let us obtain the masses of Al(OH)3 and HCl that reacted together according to the equation. This can be achieved as shown below:

Molar Mass of Al(OH)3 = 27 + 3(16+1)

= 27 + 3(17) = 27 + 51 = 78g/mol.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl from the balanced equation = 3 x 36.5 = 109.5g

Now we can obtain the mass of HCl that would react with 0.5g of Al(OH)3. This can be achieved as follow:

Al(OH)3 + 3HCl —> AlCl3 + 3H2O

From the equation above,

78g of Al(OH)3 reacted with 109.5g of HCl.

Therefore, 0.5g of Al(OH)3 will react with = (0.5 x 109.5)/78 = 0.7g of HCl

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Read 2 more answers

What is the mass of 4.40 moles of magnesium chloride, MgCl2?

Darya [45]

Since the number of moles of a substance is the mass divided by the molar mass of the substance, we can just simply multiply the molar mass of magnesium chloride by the number of moles, which is 4.40 in this case.

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Be sure to answer all parts. Hydrogen iodide decomposes according to the reaction 2 HI(g) ⇌ H2(g) + I2(g) A sealed 1.50−L contai

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Answer : The concentration of $HI$ and $I_2$ at equilibrium is, 0.0158 M and 0.00302 M respectively.

Explanation :

First we have to calculate the concentration of $H_2, I_2\text{ and }HI$

$\text{Concentration of }H_2=\frac{\text{Moles of }H_2}{\text{Volume of solution}}=\frac{0.00623mol}{1.50L}=0.00415M$

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Now we have to calculate the value of equilibrium constant (K).

The given chemical reaction is:

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Initial conc. 0.0163 0.00415 0.00276

At eqm. (0.0163-2x) (0.00415+x) (0.00276+x)

As we are given:

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Concentration of $I_2$ at equilibrium = (0.00276+x) = (0.00276+0.00026) = 0.00302 M

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