Answered

Explore Westonci.ca, the premier Q&A site that helps you find precise answers to your questions, no matter the topic. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

The double replacement reaction of aqueous Li3PO4 and CaS produce a calcium phosphate precipitate and aqueous lithium sulfide. How many grams of Ca3(PO4)2 will be produced when 1.13 grams of calcium sulfide react with excess lithium phosphate.

Sagot :

thisisbenmanley
First, use the molar mass of calcium sulfide (72.143 g/mol) to find the number of moles of CaS in 1.13 grams:

[tex](1.13\ g\ CaS)(\frac{mol}{72.143\ g})=0.01566\ mol\ CaS[/tex]

Now the equation for this reaction is 
Li3PO4 + CaS --> Ca3(PO4)2 + Li2S

but the BALANCED equation is

2Li3PO4 + 3CaS --> Ca3(PO4)2 + 3Li2S

because in this, there is an equal number of every atom.  In the balanced equation, there are 3 moles of CaS for every one mole of Ca3(PO4)2 created, so we change it like this:

[tex](0.01566\ mol\ CaS)(\frac{mol\ Ca_3(PO_4)_2}{3\ mol\ CaS})=0.00522\ mol\ Ca_3(PO_4)_2[/tex]

The molar mass of Ca3(PO4)2 is 310.1767 g/mol:

[tex](0.00522\ mol\ Ca_3(PO_4)_2)(\frac{310.1767\ g}{mol})=1.619\ g\ Ca_3(PO_4)_2[/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.