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3 years ago

Blance equation CaBr2 (aq) + _Li3PO4(aq) → _Ca3(PO4)2(s) + __LiBr (aq)

Chemistry

1 answer:

QveST [7]3 years ago

4 0

Answer:

3CaBr2 + 2LI3PO4 - > Ca3(PO4) 2 + 6LiBr

Explanation:

The first one I did was PO4. There are two on the right side, so I added 2 to Li3PO4 on the other side. That balanced the PO4s and then gave me 6 Lithiums so I balanced that one next on the right side. I added 6 to LiBr which balanced the Li but then gave me 6 Br, so I finished it off by adding 3 in front of CaBr2 which balanced the calcium and bromines.

Here was the process:

CaBr2+2Li3PO4 -> Ca3(PO4)2+LiBr

Balances PO4 (2on both sides)

CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Lithiums (6 on each side)

3CaBr2+2Li3PO4 -> Ca3(PO4)2+6LiBr

Balances Calciums and Bromines (3 Calciums and 6 Bromines each side)

Hope this helped!

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2 years ago

(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 12g Zinc Solid + 24g Silver Nitrate

VARVARA [1.3K]

Answer:

Zn + 2AgNO₃ → Zn(NO₃)₂ +2Ag

13.34g Zn(NO₃)₃ and 15.24g Ag are formed if reaction is 100% complete.

Explanation:

molar mass of Zn=65 g

molar mass of AgNO₃=170 g

molar mass of Zn(NO₃)₂ =189 g

molar mass of Silver = 108 g

Zn + 2 AgNO₃ → Zn(NO₃)₂ +2 Ag eq(1)

1 mole Zn reacts with 2 moles Silver nitrate to give 1 mole Zinc nitrate and 2 moles Silver

65 g Zn reacts with 2×170 g of AgNO₃ → 189 g Zn(NO₃)₂ and 2×108 g Ag - eq(2)

First find limiting reagent of the reaction

65g Zn reacts with 2×170g of AgNO₃

12 g Zn reacts with (12÷65)×2×170 g AgNO₃

=62.7 g AgNO₃

For the reaction to go to 100% yield 12 g Zn will need 62.7 g AgNO₃

but amount of AgNO₃ is 24 g

So the reaction yields is limited by amount of AgNO₃.

AgNO₃ is the limiting reagent.

So calculate the yield of products with the amount of AgNO₃

by eq(2)

2× 170 gAgNO₃ gives 189 g Zn(NO₃)₃

=340 g AgNO₃ gives 189 g Zn(NO₃)₃

24 g AgNO₃ gives (24÷340) ×189 g Zn(NO₃)₃

=13.34g Zn(NO₃)₃

again by eq2

2×170 g AgNO₃ gives 2×108 g Ag

= 340 g AgNO₃ gives 216 g Ag

24 g AgNO₃ gives (24÷340)×216 g Ag

= 15.24g Ag

6 0

3 years ago

Blance equation __CaBr2 (aq) + ___Li3PO4(aq) → ___Ca3(PO4)2(s) + ____LiBr (aq)

Blance equation CaBr2 (aq) + _Li3PO4(aq) → _Ca3(PO4)2(s) + __LiBr (aq)