What volume of water must be added to 11.1 mL of a pH 2.0 solution of HNO3 in order to change the pH to 4.0? A) 11.1 mL B) 89 mL

1.09 moles of tin (Sn)

Explanation:

The reaction we are examining involves tin chloride (SnCl₂) reacting with sodium (Na) to yield tin (Sn) and sodium chloride (NaCl):

SnCl₄ + 4 Na → Sn + 4 NaCl

Considering the chemical equation, this results in:

If 1 mole of SnCl₂ reacts with 4 moles of Na,

then 2.34 moles of SnCl₂ will react with X moles of Na.

X = (2.34 × 4) / 1 = 9.36 moles of Na

So, 2.34 moles of SnCl₂ can react with 9.36 moles of Na, but we have only 4.36 moles of Na on hand. Therefore, Na is the limiting reactant. Based on this, we can conclude:

If 4 moles of Na yield 1 mole of Sn,

then 4.36 moles of Na will produce Y moles of Sn.

Y = (4.36 × 1) / 4 = 1.09 moles of Sn

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limiting reactant

The calculated theoretical yield of ammonia is 29.58 Kg

In the solution provided,

the mass of is 5.22 Kg, which equals 5220 g

the mass of is 31.5 Kg, equivalent to 31500 g

The molar mass of is 2 g/mole

The molar mass of is 28 g/mole

The molar mass of is 17 g/mole

First, we must calculate the moles of and .

Moles of =

Moles of =

Next, we need to find which reagent is limiting and which is excess.

The balanced equation for the reaction will be,

Based on the balanced equation, we find that

1 mole of reacts with 3 moles of

1125 moles of react with moles of

This indicates that is the limiting reagent and is in excess.

Now, we proceed to calculate the moles of ammonia.

According to the reaction, we determine that,

3 moles of will yield 2 moles of ammonia

2610 moles of will provide moles of ammonia

Finally, we will calculate the mass of the ammonia produced.

Therefore, the theoretical yield of ammonia is 29.58 Kg