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Alika [10]
2 years ago
9

Complete the balanced equation for the reaction that occurs when formic acid (hcooh) dissolves in water. please include the stat

e of matter for all chemical species.
Chemistry
1 answer:
nignag [31]2 years ago
8 0
Your balanced equation for this reaction is:

HCOOH (aq) + H2O (aq) → HCOO- (aq) + H3O+ (aq)

So from the reaction, we can see when formic acid dissolved in water so H3O+ ions will be formed.
and we can see that it is a balanced equation as:
we have H atoms = 4 on both sides of the reaction
and C atoms = 1 atom on both sides of the reaction
and O atoms = 3 atoms on both sides of the reaction
So it is our final balanced equation of the reaction.
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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
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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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