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enyata [817]
3 years ago
6

N2 + 3H2, -> 2NH3 If I have 50.6 L of N2 and excess H2, how many liters of NH3 can I produce?

Chemistry
1 answer:
Anastaziya [24]3 years ago
3 0

Answer:

C. 101.2 L

Explanation:

N2 + H2= NH3

Balancing it,

N2 + 3 H2 = 2.NH3

(1 mol) (3 mol) (2 mol)

which means

1 molecule of nitrogen reacts with 3 molecule of hydrogen to form ammonia.

Likewise,

50.6 l of nitrogen reacts with 50.6 × 3= 151.8 l of hydrogrn to form 50.6 × 2= 101.2 l of ammonia.

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How many moles of aqueous potassium ions and sulfate ions are formed when 63.7 g of K2SO4 dissolves in water
Viefleur [7K]

Answer:

WHEN 63.7 g OF K2SO4 IS DISSOLVED IN WATER, 0.73 MOLES OF POTASSIUM ION AND 0.366 MOLES OF SULFATE ION ARE FORMED.

Explanation:

Equation for the reaction:

K2SO4 + H20 ------->2 K+ + SO4^2-

When K2SO4 dissolves in water, potassium ion and sulfate ion are formed.

1 mole of K2SO4 produces 2 moles and 1 mole of SO4^2-

At STP, 1 mole of K2SO4 will be the molar mass of the substance

Molar mass of K2SO4 = ( 39 *2 + 32 + 16*4) g/mol

Molar mass = 174 g/mol

So therefore;

1 mole of K2SO4 contains 174 g and it produces 2 moles of potassium and 1 mole of sulfate ion

When 63.7 g is used; we have:

174 g = 2 moles of K+

63.7 g = ( 63.7 * 2 / 174) moles of K+

= 0.73 moles of K+

Forr sulfate ion, we have:

174 g = 1 mole ofSO4^2-

63.7 g = (63.7 * 1 / 174) moles of SO4^2-

= 0.366 moles of SO4^2-

In other words, when 63.7 g of K2SO4 is dissolved in water, 0.73 moles of potassium ion and 0.366 moles of sulfate ion are formed.

4 0
3 years ago
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