C3H8 +502 + 3CO2 + 4H2O
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The question is incomplete. Here is the complete question.
A nitric acid solution containing 71.0% HNO₃ (by mass) has a density of 1.42g/mL. How many moles of HNO₃ are present in 1.98L of this solution?
a. 62.8 mol
b. 31.7 mol
c. 22.3 mol
d. 3.16 x 10⁻² mol
e. none of these
Answer: b. 31.7 mol
Explanation: The solution has density of 1.42g/mL. So, in 1980 mL of solution:
m = d.V
m = 1.42(1980)
m = 2811.6 g
There are 2811.6 grams of solute.
Nitric acid in this solution is 71% by mass, which means
m = 2811.6(0.71)
m = 1996.236 g
In this solution, mass of nitric acid is 1996.236 grams.
Molar mass of HNO₃ is M = 63.01g/mol.
Then:
n = 31.68 moles
In 1.98L of a nitric acid solution, there are 31.7 moles of HNO₃.