I will give brainiest answer to whoever helps me, also this problem is worth 20 points
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The question is incomplete, here is the complete question:
Assume that concentrated aqueous NH₃ has a density of 0.252 g/mL (0.252 g of NH₃ per mL of liquid). Calculate the volume of NH₃ required to contain 0.442 mol?
<u>Answer:</u> The volume of ammonia required is 29.82 mL
<u>Explanation:</u>
To calculate the number of moles, we use the equation:
Molar mass of ammonia = 17 g/mol
Moles of ammonia = 0.442 moles
Putting values in above equation, we get:
To calculate the volume of ammonia, we use the equation:
Density of ammonia = 0.252 g/mL
Mass of ammonia = 7.514 g
Putting values in above equation, we get:
Hence, the volume of ammonia required is 29.82 mL
Answer: 2.796 grams
Explanation:
According to stoichiometry:
1 mole of reacts with 1 mole of
0.012 moles of will react with=
of
Thus is the limiting reagent as it limits the formation of product.
is the excess reagent as (0.018-0.012)=0.006 moles are left unused.
1 mole of produces 1 mole of
0.012 moles of will produce=
of
Mass of
Thus 2.796 grams of are produced.