How many moles of chromium (lll) nitrate are produced when chromium reacts with 0.85 moles of lead (IV) Nitrate to produce chromium (lll) nitrate and lead?

Question

asked May 26, 2021 184k views

1 Answer

Tom Bevelander · Answer 1 · 2021-05-30T14:25:30+0000

Answer:

1.133 moles of chromium (lll) nitrate are produced.

Step-by-step explanation:

$4Cr+3Pb(NO_3)_4\rightarrow 4Cr(NO_3)_3+3Pb$

Moles of lead(IV) nitrate = 0.85 mole

According to recation, 3 moles of lead(IV) nitrate gives 4 moles of chromium (III) nitrate.

Then 0.85 moles of lead(IV) nitrate will give:

(4)/(3)* 0.85 mol=1.133 mol of chromium (lll) nitrate.

1.133 moles of chromium (lll) nitrate are produced.