Question

M(Al(NO3)2=42,6 гm(NaOH)=16 гm(AlOH3)-?

Answer 1

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Answer 2

To find the mass of Al(OH)3 formed when 42.6 grams of Al(NO3)2 reacts with 16 grams of NaOH, we need to set up and balance the chemical equation for the reaction between Al(NO3)2 and NaOH. The reaction can be represented as follows:

2 Al(NO3)2 + 6 NaOH → 2 Al(OH)3 + 3 NaNO3

From the balanced equation, we can see that 2 moles of Al(NO3)2 react with 6 moles of NaOH to produce 2 moles of Al(OH)3.

Now, let's find the molar masses of the involved compounds:

Molar mass of Al(NO3)2: Al = 26.98 g/mol N = 14.01 g/mol O = 16.00 g/mol Molar mass of Al(NO3)2 = 26.98 + (14.01 * 2) + (16.00 * 6) = 26.98 + 28.02 + 96.00 = 150 g/mol

Molar mass of NaOH: Na = 22.99 g/mol O = 16.00 g/mol H = 1.01 g/mol Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol

Now, we can calculate the number of moles for each reactant:

Number of moles of Al(NO3)2 = mass / molar mass = 42.6 g / 150 g/mol ≈ 0.284 moles

Number of moles of NaOH = mass / molar mass = 16 g / 40 g/mol = 0.4 moles

Since the reaction ratio is 2 moles of Al(NO3)2 to 2 moles of Al(OH)3, we need to use the stoichiometry to find the number of moles of Al(OH)3 formed. From the balanced equation, we see that 2 moles of Al(NO3)2 produce 2 moles of Al(OH)3.

So, the number of moles of Al(OH)3 formed = 0.284 moles (moles of Al(NO3)2) * (2 moles Al(OH)3 / 2 moles Al(NO3)2) = 0.284 moles

Now, calculate the mass of Al(OH)3 formed:

Mass of Al(OH)3 = number of moles * molar mass = 0.284 moles * (26.98 + (1.01 * 3) + (16.00 * 3)) ≈ 0.284 moles * 78.03 g/mol ≈ 22.18 grams

So, approximately 22.18 grams of Al(OH)3 will be formed in the reaction.