Question

M(AlCl3)=25 г Найти m(Naoh) AlCl3+NaOH=....

Answer 1

дано

m(AlCL3) = 25 g

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m(NaOH)-?

AlCL3+3NaOH-->Al(OH)3+3NaCL

M(AlCL3) = 133.5 g/mol

n(AlCL3) = m/M = 25 / 133.5 = 0.19 mol

n(AlCl3) = 3n(NaOH)

n(NaOH) = 3*0.19 = 0.57 mol

M(NaOH) = 40 g/mol

m(NaOH) = n*M= 0.57 * 40 = 22.8 g

ответ 22.8 г

Answer 2

To find the mass of NaOH (sodium hydroxide) required to react with 25 grams of AlCl3 (aluminum chloride), we need to determine the balanced chemical equation for the reaction between AlCl3 and NaOH. The balanced chemical equation is as follows:

2AlCl3 + 6NaOH → 2Al(OH)3 + 6NaCl

From the balanced equation, we can see that 2 moles of AlCl3 react with 6 moles of NaOH to produce 2 moles of Al(OH)3 and 6 moles of NaCl.

Now, we need to calculate the molar masses of the compounds involved:

Molar mass of AlCl3 = 26.98 g/mol (Al) + 3 * 35.45 g/mol (Cl) = 133.33 g/mol Molar mass of NaOH = 22.99 g/mol (Na) + 15.999 g/mol (O) + 1.008 g/mol (H) = 39.997 g/mol ≈ 40 g/mol

Next, we convert the given mass of AlCl3 into moles:

moles of AlCl3 = mass / molar mass moles of AlCl3 = 25 g / 133.33 g/mol ≈ 0.1875 moles

Now, we can use the mole ratio from the balanced equation to find the moles of NaOH required:

moles of NaOH = moles of AlCl3 * (6 moles NaOH / 2 moles AlCl3) moles of NaOH = 0.1875 moles * (6 / 2) ≈ 0.5625 moles

Finally, we convert the moles of NaOH back into mass:

mass of NaOH = moles of NaOH * molar mass of NaOH mass of NaOH = 0.5625 moles * 40 g/mol ≈ 22.5 grams

So, approximately 22.5 grams of NaOH are required to react completely with 25 grams of AlCl3.