Question

5
DU TU, V - ID, L 12 dd L- 5.) A
2.0 g sample containing Na, CO and NaHCO, loses 0.248 when heated to 300°C, the tempera-
ture at which NaHCO, decomposes to Na CO,, and H,0. What is% of Na,Co, in mixture?
2NaHCO, - Na,CO, + CO, + H0
​

Answer 1

Explanation:

Answer: The volume of nitrogen monoxide produced is 12 L.

Explanation:

For the given chemical reaction:

4NH_3+5O_2\rightarrow 4NO+6H_2O4NH

3

+5O

2

→4NO+6H

2

O

According to the mole concept:

1 mole of a gas occupies 22.4 L of volume.

So, 9 moles of reactants (which are gases) will occupy 9\times 22.4=201.6L9×22.4=201.6L of volume.

By Stoichiometry of the reaction:

201.6 L of volume of reactants produces (4\times 22.4)=89.6L(4×22.4)=89.6L of nitrogen monoxide.

So, 27 L of reactants will produce = \frac{89.6}{201.6}\times 27=12L

201.6

89.6

×27=12L of nitrogen monoxide.

Thus, the volume of nitrogen monoxide produced is 12 L.

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Answer 2

Answer:

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