Al2O3 would be typed in as Al2O3 Al2(SO4)3 would be typed in as Al2(SO4)3. Required:
a. Write the formula for calciumnitride.
b. Write the formula for magnesium phosphide
c. Write the formula for rubidium chromate
d. Write the formula for aluminum nitrate
e. Write the formula for ammonium arsenide
f. Write the formula for nickel(II) nitrite
g. Write the formula for copper(I) sulfate
h. Write the formula for iron(III) nitrate
i. Write the formula for manganese(II) nitride

Explanation:

Hello,

We are required to write the chemical formula of the following compounds

1. Calcium nitride = Ca₃N₂

2. Magnesium Phosphide = Mg₃P₂

3. Rubidium Chromate = Rb₂CrO₄

4. Aluminium nitrate = Al(NO₃)₃

5. Ammonium Arsenide = (NH₄)₃As

6. Nickel(ii) nitrite = Ni(NO₂)₂

7. Copper(i)sulfate = Cu₂SO₄

8. Iron(iii)nitrate = Fe(NO₃)₃

9. Manganese(ii)nitrate = Mn(NO₃)₂

The left hand side are the common names when the right hand side are the chemical formula.

Sodium hydroxide (NaOH) is the limiting reactant

Explanation:

Step 1: Data given

Number of moles of NaOH = 15.0 mol

Number of moles of H3PO4 = 7.50 mol

Molar mass of NaOH = 40 g/mol

Molar mass of H3PO4 = 98 g/mol

Step 2: The balanced equation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

Step 3: Calculate the limiting reactant

For 1 mole of H3PO4, we need 3 moles of NaOH

NaOH is the limiting reactant. It will completely be consumed. (15 moles)

H3PO4 is in excess, there will react 15.0 / 3 = 5 moles

There will remain 7.5 - 5.0 = 2.5 moles

There will be produced 15.0/3 = 5.0 moles of Na3PO4

There will be produced 15.0 moles of H2O

Sodium hydroxide (NaOH) is the limiting reactant