Answer:
1.133 moles of chromium (lll) nitrate are produced.
Explanation:
[tex]4Cr+3Pb(NO_3)_4\rightarrow 4Cr(NO_3)_3+3Pb[/tex]
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:
[tex]\frac{4}{3}\times 0.85 mol=1.133 mol[/tex] of chromium (lll) nitrate.
1.133 moles of chromium (lll) nitrate are produced.
