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melomori [17]
3 years ago
7

Sodium hydride reacts with excess water to produce aqueous sodium hydroxide and hydrogen gas:NaH (s) H2O (l) → NaOH (aq) H2 (g)A

sample of NaH weighing ________ g will produce 982 mL of gas at 28.0 °C and 765 torr, when the hydrogen is collected over water. The vapor pressure of water at this temperature is 28 torr.
Chemistry
2 answers:
Anuta_ua [19.1K]3 years ago
7 0

NaH(s)+ H2O (l)=>NaOH(aq)+H2(g)

You want to calculate the mass of NaH, I assume.  Otherwise, the question isn't clear.  It simply says calculate the mass(??)

 

So, calculate the moles of H2 gas that satisfy the conditions of 982 ml at 28ºC and 765 torr.  But you must subtract the vapor pressure of water at 28º to get the actual pressure of the H2 gas.  So, the actual conditions are 982 ml (0.982 L) and 301 K and 765-28 = 737 torr.

PV = nRT

n = PV/RT = (737 torr)(0.982 L)/(62.4 L-torr/Kmol)(301 K)

n = 0.0385 moles H2

 

moles NaH needed = 0.0385 moles H2 x 1 mole NaH/mole H2 = 0.0385 moles NaH required

mass of NaH needed = 0.0385 moles x 24 g/mole = 0.925 g NaH

Brainliest Please :)

gizmo_the_mogwai [7]3 years ago
7 0

Answer:

530

Explanation:

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6 0
3 years ago
Suppose you are titrating vinegar, which is an acetic acid solution
kotegsom [21]

Answer:

0.373 M

Explanation:

The balanced equation for the reaction is given below:

HC2H3O2 + NaOH —> NaC2H3O2 + H2O

From the balanced equation above, the following were obtained:

Mole ratio of the acid, HC2H3O2 (nA) = 1

Mole ratio of the base, NaOH (nB) = 1

Next, we shall write out the data obtained from the question. This include:

Volume of base, NaOH (Vb) = 32.17 mL

Molarity of base, NaOH (Mb) = 0.116 M

Volume of acid, HC2H3O2 (Va) = 10 mL

Molarity of acid, HC2H3O2 (Ma) =..?

The molarity of the acid solution can be obtained as follow:

MaVa/MbVb = nA/nB

Ma x 10 / 0.116 x 32.17 = 1

Cross multiply

Ma x 10 = 0.116 x 32.17

Divide both side by 10

Ma = (0.116 x 32.17) /10

Ma = 0.373 M

Therefore, the concentration of the acetic acid is 0.373 M.

8 0
3 years ago
the combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how many moles of water formed
enyata [817]
2C_4H_{10}+13O_2 ⇒ 8CO_2 + 10H_2O

n= \frac{m}{M} =  \frac{m}{2M(H)+M(O)}= \frac{2,46}{2*1,0+16,0}  = 0,14mol

So 0,14mol are formed.


6 0
3 years ago
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