In the following reaction: Au(NO3)3 (aq) + 3 Li(s) ➞ Au(s) + 3 LiNO3 (aq)a. How many grams of Lithium would I need to completely react 24g if gold nitrate?molar mass of Au(NO3)3: 382.97 g/molmolar mass of LiNO3: 68.95 g/molmolar mass of Li: 7 g/molb. How much gold, in grams, would you produce from the reaction? (Molar mass of Au: 196.97 g/mol)

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Asked Nov 12, 2019
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In the following reaction: Au(NO3)3 (aq) + 3 Li(s) ➞ Au(s) + 3 LiNO3 (aq)

a. How many grams of Lithium would I need to completely react 24g if gold nitrate?

molar mass of Au(NO3)3: 382.97 g/mol

molar mass of LiNO3: 68.95 g/mol

molar mass of Li: 7 g/mol

b. How much gold, in grams, would you produce from the reaction? (Molar mass of Au: 196.97 g/mol)

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Expert Answer

Step 1

(a)

Given that 24 g of Au(NO3)3 is present and the balanced equation is given below:

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Au (NO,), (aq)+3Li(s)- Au(s)+ 3LiNO, (aq) 1 molAu (NO)requires ->3molLi 24g No.of moles of Au(NO;) : molarmass 24.0g 382.97g/mol 0.0626mol No.of moles of Li=3x0.0626 mol 0.1878mol

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Step 2

From the number of moles and molecular mass...

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mass No.of moles molar mass mass No.of molesxmolarmass Mass of Li-0.1878 molx7 g/mol Mass of Li-1.315g

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