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Neon belongs to 2nd Period & 18th Group
Answer:
THE VOLUME OF 0.200M CALCIUM HYDROXIDE NEEDED TO NEUTRALIZE 35 mL of 0.050 M NITRIC ACID IS 43.75 mL.
Explanation:
Using
Ca VA / Cb Vb = Na / Nb
Ca = 0.0500 M
Va = 35 mL
Cb = 0.0200 M
Vb = unknown
Na = 2
Nb = 1
Equation for the reaction:
Ca(OH)2 + 2HNO3 --------> Ca(NO3)2 + 2H2O
So therefore, we make Vb the subject of the equation and solve for it
Vb = Ca Va Nb / Cb Na
Vb = 0.0500 * 35 * 1 / 0.0200 * 2
Vb = 1.75 / 0.04
Vb = 43.75 mL
The volume of 0.02M calcium hydroxide required to neutralize 35 mL of 0.05 M nitric acid is 43.75 mL
The number of moles of NH3 that could be made would be 0.5 moles
<h3>Stoichiometric reactions</h3>
From the balanced equation of the reaction:
N2 (g) + 3 H2(g) ----> 2NH3 (g)
The mole ratio of N2 to H2 is 1:3
Thus, for 0.50 moles of N2, 1.5 moles of H2 should be present. But 0.75 moles of H2 was allowed to react. Meaning that H2 is limiting in this case.
Mole ratio of H2 and NH3 = 3:2
Thus for 0.75 moles H2, the mole of NH3 that would be produced will be:
2 x 0.75/3 = 0.5 moles
More on stoichiometric calculations can be found here: brainly.com/question/8062886
Answer:
Yes
Explanation:
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