It says to to use the word sub to indicate the subscript. for example, o2 would be written as osub2. 1. complete and balance the the following equation: cd(no3)2+nh4cl > 2. write a balanced net ionic equation for the following reaction. h3po4(aq)+ca(oh)2(aq) > ca3 (po4)2(aq)+h2o(l) show all work . 

Explanation:

Question 1.

The products in this chemical reaction are; CdCl2 and  NH4NO3.

So we have;

Cd(NO₃)₂ + NH₄Cl ---> CdCl₂ + NH₄NO₃

The reaction is not balanced because we have unequal number of atoms of elements in both sides of the equation.

Upon balancing, we now have;

Cd(NO₃)₂ + 2NH₄Cl ---> CdCl₂ + 2NH₄NO₃

Question 2.

To write a net ionic equation:

- Write the balanced molecular equation.

2H₃PO₄(aq) + 3Ca(OH)₂(aq) ---> Ca₃ (PO₄)₂(aq) + 6H₂O(l)

- Write the balanced complete ionic equation.

6H+(aq] + 2PO³⁻₄(aq] + 3Ca²⁺(aq] + 6OH⁻(aq]  → 3Ca²⁺(aq] + 2PO³⁻₄(aq] + 6H₂O(l]

Cross out the spectator ions that are present and Write the "leftovers" as the net ionic equation.

You can recognize spectator ions by looking for ions that are present on both sides of the equation. They will always have the same exact formula, charge, and physical state. They will also be present in exactly the same number on both sides of the equation.

6H⁺(aq] + 6OH⁻(aq] --> 6H₂O(l]

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

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