Answered

Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

Can someone help me with these stoichiometry problems please!?

Can Someone Help Me With These Stoichiometry Problems Please class=

Sagot :

pstnonsonjoku

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

Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.