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Can someone help me with these stoichiometry problems please!?

Can Someone Help Me With These Stoichiometry Problems Please class=

Sagot :

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Answer:

See explanation

Explanation:

The equation of the reaction is;

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

a) From the reaction equation;

2 moles of H3PO4 reacts  with 3 moles of Ca(OH)2

16.5 moles of H3PO4 reacts  with 16.5 * 3/2 = 24.75 moles of Ca(OH)2

Mass = 24.75 moles of Ca(OH)2 * 74.093 g/mol

Mass = 1833.8 g of Ca(OH)2

b) if 1 mole of Ca(OH)2 = 6.02 * 10^23 molecules

x moles of Ca(OH)2 = 5.06 * 10^24 molecules

x = 5.06 * 10^24 molecules * 1 mole/6.02 * 10^23 molecules

x = 8.4 moles

3 moles of Ca(OH)2 produces 6 moles of H2O

8.4 moles of Ca(OH)2 yields x moles of H2O

x = 8.4 * 6/3

x = 16.8 moles of H20

1 mole of H2O = 6.02 * 10^23 molecules

16.8 moles of H20 =  16.8 moles * 6.02 * 10^23 molecules/ 1 mole

= 1.01 * 10^25 molecules of H2O

c) since 1 mole of Ca3(PO4)2 occupies 22.4 L and only 1 mole of Ca3(PO4)2 is produced in the reaction so;

22.4 L is occupied by  Ca3(PO4)2 when 6(22.4) L of water is produced

x litres of  Ca3(PO4)2 is occupied when 4.2 L of water is produced

x = 22.4 L * 4.2 L/ 6(22.4) L

x = 0.7 L of Ca3(PO4)2

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