Answer all the questions in the space provided. Balance the following equations:

Question

Answer all the questions in the space provided. Balance the following equations: i). Na_2 O(s) + H_2 O(l) rightarrow NaOH(aq): Ans. _ ii) MgCO_3 (s) + HNO_3 (aq) rightarrow Mg(NO_3)_2+ CO_2 (g) + H_2 O(1) Ans: _ iii) NH_3 (aq) + H_2 SO_4 (aq) rightarrow (NH_4 SO-4 (aq) Ans. _ iv) H_3 PO_4 (aq) + NaOH(aq) rightarrow Na_2 HPO_4 (aq) + H_2 O(l) Ans: _ Calculate the volume of 0.1005 M HC1 that would be needed to neutralize exactly 0.2505 g of MgO according to the reaction: MgO(s) + 2 HCl(aq) rightarrow MgCl_2 (aq) + H_2 O(l) Ans. _ A 0.2076 g of Na_2 CO_3 required 20.35 mL of a hydrochloric acid solution for complete neutralization: Na_2 CO_3 (aq) + 2 HCl(aq) rightarrow 2 NaCl(aq) + CO_2 (g) + H_2 O(1) What is the molarity of the HC1 solution? Ans. _

Explanation / Answer

1) i) Na2O + H2O - 2NaOH

ii) Metal Carbonate + Acid Salt + Carbon Dioxide + Hydrogen

MgCO3(s) + 2HNO3(aq) Mg(NO3)2(aq) + CO2(g) + H2O(l)

It's Mg(NO3)2 (and not MgNO3) because the magnesium ion has a charge of +2, and so you need two nitrate ions to equalize the charge (because NO3 ions only have a charge of -1).

iii)  2 NH3 + H2SO4 --> (NH4)2SO4

iv)   2NaOH(aq) + H3PO4(aq) --> Na2HPO4(aq) + 2H2O(l)

2) Assume 10 g solution 0.2505 g is MgO

100g/0.2505g =399.20 ml =3.9920 L

moles =concentration x volume

= 0.1005 M x 3.9920L

=0.40119 moles

volume=moles/ concentration

=0.40119 / 0.1005

=4.00 ml

3) Molarity = Moles of solute / Kg of solvent

Moles of Hcl = Mass/ Molar mass

Molar mass of Hcl =36.5 g/mol

Assume 100 g of solution 10 g is Hcl

10 g Hcl / 36.5 g/mol = 0.273 moles Hcl

100g/20.35 ml = 49.14ml= 0.04914 L

Assume 1000g solution

Molarity= 0.273 moes/ 0.004914

=55.55 M

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