Write the concentrated sulphuric acid H2C2O42H2O=2H2O+CO2+CO
1 answer:
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Answer:
3.7 g
Explanation:
In this problem, marble reacted <em>completely</em> with an excess of HCl, which then reacted with NaOH
The reaction between marble and HCl is:
CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
Then the excess HCl reacted with NaOH:
HCl + NaOH → NaCl + H₂O
So first let's <u>calculate the amount of excess HCl that reacted with NaOH:</u>
0.9987 M NaOH * 0.02487 L * = 0.02483 mol HCl
Then we calculate the number of unreacted HCl moles in the original 2 L:
0.02483 mol HCl * 2 L / 0.01 L = 4.966 mol HCl
We <u>substract that number from the initial HCl moles</u> in order to find out how many of them reacted with the piece of marble
Initial HCl moles = 2.00 L * 2.52 M = 5.04 mol HCl
HCl moles that reacted with marble = 5.04 - 4.966 = 0.074 mol HCl
With that we can <u>calculate the moles of marble</u>, and <u>finally the mass</u>, using its molecular weight:
0.074 mol HCl * *
= 3.7 g marble