Question

Can someone please give and help. I have a hard time with stoich. Pls explain

Answer 1

2) The mass of the ammonium hydroxide 33.6 g.

3) The mass of the calcium hydroxide is 5.18 g.

2) The balanced reaction equation is;

`BeSO_4 + 2NH_4OH --- > Be(OH)_2 + (NH_4)_2SO_4`

Number of moles of
`BeSO_4` = 50 g/105 g/mol

= 0.48 moles

If 1 moles of
`BeSO_4` reacts with 2 moles of ammonium hydroxide

0.48 moles of
`BeSO_4` reacts with 0.48 * 2/1

= 0.96 moles

Mass of the ammonium hydroxide = 0.96 moles * 35 g/mol

= 33.6 g

3) The reaction equation is;

`2Ca(OH)_2 + P_4O_(10) + 2H_2O --- > 2Ca(H_2PO_4)_2`

Number of moles of phosphorus oxide = 10 g/284 g/mol

= 0.035 moles

2 moles of calcium hydroxide reacts with 1 of phosphorus oxide

x moles of calcium hydroxide reacts with 0.035 moles of phosphorus oxide

x = 2 * 0.035 /1

= 0.07 moles

Mass of the calcium hydroxide

= 0.07 moles * 74 g/mol

= 5.18 g