Question

How many moles of Chromium(III) nitrate (Cr(NO3)3) are produced when Chromium reacts with 0.85 moles of Lead(IV) nitrate (Pb(NO3)4) to produce chromium(III) nitrate (Cr(NO3)3) and lead? 4 Cr + 3 Pb(NO3)4 → 4 Cr(NO3)3 + 3 Pb

Answer 1

Answer:

$0.64molCr(NO_3)_3$

Explanation:

Hello there!

In this case, according to the given chemical equation, it turns out possible for us to realize there is a 4:3 mole ratio of chromium(III) nitrate to lead(IV) nitrate, and therefore, we can calculate the moles of the former by applying the shown below stoichiometry setup:

$0.85molPb(NO_3)_4*\frac{3molCr(NO_3)_3}{4molPb(NO_3)_4} \\\\0.64molCr(NO_3)_3$

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