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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?

Respuesta :

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.

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