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sineoko [7]
3 years ago
15

How many moles of KOH are required to produce 4.79 g K3PO4 according to the following reaction? 3KOH + H3PO4 -----> K3PO4 + 3

H2O
Chemistry
1 answer:
8_murik_8 [283]3 years ago
5 0

Answer:

0.677 moles

Explanation:

Take the atomic mass of K = 39.1, O =16.0, P = 31.0

no. of moles = mass / molar mass

no. of moles of K3PO4 used = 4.79 / (39.1x3 + 31 + 16x4)

= 0.02256 mol

From the equation, the mole ratio of KOH : K3PO4 = 3 :1,

meaning every 3 moles of KOH used, produces 1 mole of K3PO4.

So, using this ratio, let the no. of moles of KOH required to be y.

\frac{3}{1} =\frac{y}{0.02256} \\

y = 0.02256 x3

y = 0.0677 mol

If you don't find exactly 0.677 moles as one of the options, go for the closest one. A very slight error may occur because of taking different significant figures of atomic masses when calculating.

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2 years ago
how many kilograms of a 35% m/m sodium chlorate solution is needed to react completely with 0.29 l of a 22% m/v aluminum nitrate
Stolb23 [73]

Answer:- 0.273 kg

Solution:- A double replacement reaction takes place. The balanced equation is:

3NaClO_3+Al(NO_3)_3\rightarrow 3NaNO_3+Al(ClO_3)_3

We have 0.29 L of 22% m/v aluminum nitrate solution. m/s stands for mass by volume. 22% m/v aluminium nitrate solution means 22 g of it are present in 100 mL solution. With this information, we can calculate the grams of aluminum nitrate present in 0.29 L.

0.29L(\frac{1000mL}{1L})(\frac{22g}{100mL})

= 63.8 g aluminum nitrate

From balanced equation, there is 1:3 mol ratio between aluminum nitrate and sodium chlorate. We will convert grams of aluminum nitrate to moles and then on multiplying it by mol ratio we get the moles of sodium chlorate that could further be converted to grams.

We need molar masses for the calculations, Molar mass of sodium chlorate is 106.44 gram per mole and molar mass of aluminum nitrate is 212.99 gram per mole.

63.8gAl(NO_3)_3(\frac{1mol}{212.99g})(\frac{3molNaClO_3}{1molAl(NO_3)_3})(\frac{106.44g}{1mol})

= 95.7gNaClO_3

sodium chlorate solution is 35% m/m. This means 35 g of sodium chlorate are present in 100 g solution. From here, we can calculate the mass of the solution that will contain 95.7 g of sodium chlorate  and then the grams are converted to kg.

95.7gNaClO_3(\frac{100gSolution}{35gNaClO_3})(\frac{1kg}{1000g})

= 0.273 kg

So, 0.273 kg of 35% m/m sodium chlorate solution are required.

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3 years ago
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Read 2 more answers
Calculate the mole fraction of the ionic species kcl in the solution.
sp2606 [1]

The question is incomplete, here is the complete question:

Calculate the mole fraction of the ionic species KCl in the solution A solution was prepared by dissolving 43.0 g of KCl in 225 g of water.

<u>Answer:</u> The mole fraction of KCl in the solution is 0.044

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}      .....(1)

  • <u>For water:</u>

Given mass of water = 225 g

Molar mass of water = 18 g/mol

Putting values in equation 1, we get:

\text{Moles of water}=\frac{225g}{18g/mol}=12.5mol

  • <u>For KCl:</u>

Given mass of KCl = 43 g

Molar mass of KCl = 74.55 g/mol

Putting values in equation 1, we get:

\text{Moles of KCl}=\frac{43g}{74.55g/mol}=0.577mol

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

  • <u>For KCl:</u>

Moles of KCl = 0.577 moles

Total moles = [0.577 + 12.5] = 13.077 moles

Putting values in above equation, we get:

\chi_{(KCl)}=\frac{0.577}{13.077}=0.044

Hence, the mole fraction of KCl in the solution is 0.044

8 0
2 years ago
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