Question

Натрий-Гидроксид массой 4,0 г сплавили алюминием-гидроксидом массой 9,8 г . Рассчитайте массу образовавшегося натрий-метаалюмината.

Answer 1

уравнение реакции:

NaOH + Al(OH)3 = NaAlO2 + 2H2O

Количества взятых веществ:

n(NaOH) = m(NaOH)/M(NaOH) = 4/40 = 0,1 моль

n(Al(OH)3) = 9,8/78 = 0,13 моль

- гидроксид алюминия взят в избытке, количество образующегося метаалюмината будет равно количеству взятого гидроксида натрия, тогда его масса

m(NaAlO2) = n(NaAlO2)*M(NaAlO2) = 0,1*82 = 8,2 г

Answer 2

2NaOH + 2Al(OH)3 ---> 2NaAlO2 + 4H2O

кол-во в-ва(NaOH)= 4/40*2=0,05 моль - недостатке, решаем по нему

кол-во в-ва(Al(OH)3)=9,8/156= 0,063 моль

4----------x

80--------164

x=164*4/80=8,2 г

 

Answer 3

Calculation of the Mass of Sodium Metaaluminate

To calculate the mass of sodium metaaluminate formed when 4.0 g of sodium hydroxide reacts with 9.8 g of aluminum hydroxide, we need to determine the balanced chemical equation for the reaction and use stoichiometry to find the mass of the product.

The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and aluminum hydroxide (Al(OH)3) is:

2NaOH + Al(OH)3 → Na2Al2O4 + 3H2O

From the balanced equation, we can see that 2 moles of sodium hydroxide react with 1 mole of aluminum hydroxide to produce 1 mole of sodium metaaluminate.

To calculate the mass of sodium metaaluminate, we can follow these steps:

Step 1: Convert the given masses of sodium hydroxide and aluminum hydroxide to moles using their respective molar masses.

The molar mass of sodium hydroxide (NaOH) is: Na: 22.99 g/mol O: 16.00 g/mol H: 1.01 g/mol

So, the molar mass of NaOH is: 22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol

The number of moles of NaOH is: 4.0 g / 40.00 g/mol = 0.10 mol

The molar mass of aluminum hydroxide (Al(OH)3) is: Al: 26.98 g/mol O: 16.00 g/mol H: 1.01 g/mol

So, the molar mass of Al(OH)3 is: 26.98 g/mol + (16.00 g/mol × 3) + (1.01 g/mol × 3) = 78.00 g/mol

The number of moles of Al(OH)3 is: 9.8 g / 78.00 g/mol = 0.13 mol

Step 2: Determine the limiting reactant.

To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric coefficients in the balanced equation.

From the balanced equation, we can see that the stoichiometric ratio between NaOH and Al(OH)3 is 2:1. This means that 2 moles of NaOH react with 1 mole of Al(OH)3.

Since we have 0.10 mol of NaOH and 0.13 mol of Al(OH)3, we can see that NaOH is the limiting reactant because it is present in a smaller amount.

Step 3: Calculate the moles of sodium metaaluminate formed.

From the balanced equation, we know that 2 moles of NaOH react to form 1 mole of sodium metaaluminate (Na2Al2O4).

Since NaOH is the limiting reactant, we can calculate the moles of sodium metaaluminate formed using the stoichiometric ratio:

0.10 mol NaOH × (1 mol Na2Al2O4 / 2 mol NaOH) = 0.05 mol Na2Al2O4

Step 4: Convert the moles of sodium metaaluminate to grams using its molar mass.

The molar mass of sodium metaaluminate (Na2Al2O4) is: Na: 22.99 g/mol Al: 26.98 g/mol O: 16.00 g/mol

So, the molar mass of Na2Al2O4 is: (22.99 g/mol × 2) + (26.98 g/mol × 2) + (16.00 g/mol × 4) = 186.94 g/mol

The mass of sodium metaaluminate formed is: 0.05 mol Na2Al2O4 × 186.94 g/mol = 9.35 g

Therefore, the mass of sodium metaaluminate formed is approximately 9.35 grams.