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Romashka-Z-Leto [24]
2 years ago
5

Balance the following equation __N2+__O2→__N2O​

Chemistry
1 answer:
fenix001 [56]2 years ago
3 0

Answer:

2N₂   +  O₂  →  2N₂O

Explanation:

Given equation:

        N₂   +  O₂  →  N₂O

In balancing a chemical equation, the number of atoms on both sides of the expression must be the same.

To solve this problem, we use a simple mathematical approach;

  put coefficients a,b and c

             aN₂   +  bO₂  →  cN₂O  

Conserving N;

                   2a  = 2c

Conserving O:

                  2b  = c

 Let a = 1, c = 1 , b  = \frac{1}{2}

   multiply through by 4

           a = 2, b  = 1 and c  = 2

 Balanced equation is;

          2N₂   +  O₂  →  2N₂O

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20.0 g of bromic acid, HBrO3, is reacted with excess HBr.
Blababa [14]

After Rounding off The percentage yield is 64%

<h3>What is Percentage Yield ?</h3>

It is the ratio of actual yield to theoretical yield multiplied by 100% .

It is given in the question

20.0 g of bromic acid, HBrO3, is reacted with excess HBr.

The reaction is

HBrO₃ (aq) + 5 HBr (aq) → 3 H₂O (l) + 3 Br₂ (aq)

Actual yield = 47.3 grams

Molecular weight of Bromic Acid is 128.91 gram

Moles of Bromic Acid = 20/128.91 = 0.155 mole

Mole fraction ratio of Bromic Acid to Bromine is 1 :3

Therefore for 0.155 mole of Bromic Acid 3 * 0.155 = 0.465 mole of Bromine is produced.

1 mole of Bromine = 159.8 grams of Bromine

0.465 of Bromine = 74.31 grams of Bromine

Percentage Yield = (47.3/74.31)*100 = 63.65 %

After Rounding off The percentage yield is 64% .

To know more about Percentage Yield

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5 0
1 year ago
PLEASE HELP
swat32

Answer:

1461.7 g of AgI

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaI₂ + 2AgNO₃ —> 2AgI + Ca(NO₃)₂

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Next, we shall determine the number of mole AgI produced by the reaction of 3.11 moles of CaI₂. This can be obtained as follow:

From the balanced equation above,

1 mole of CaI₂ reacted to produce 2 moles of AgI.

Therefore, 3.11 moles of CaI₂ will react to produce = 3.11 × 2 = 6.22 moles of AgI

Finally, we shall determine the mass of 6.22 moles of AgI. This can be obtained as follow:

Mole of AgI = 6.22 moles

Molar mass of AgI = 108 + 127

= 235 g/mol

Mass of AgI =?

Mass = mole × molar mass

Mass of AgI = 6.22 × 235

Mass of AgI = 1461.7 g

Therefore, 1461.7 g of AgI were obtained from the reaction.

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