How many moles of chromium (iii) nitrate are produced when chromium reacts with 0.85 moles of lead (iv) nitrate to produce chromium (iii) nitrate and lead? word equation: formula equation: (must be balanced! )

1.13 mol

Explanation:

1) Word equation:

chromium + lead(IV) nitrate → chromium(III) nitrate + lead

2) Chemical formula of the compounds (salts)

lead(IV) nitrate: Pb (NO₃)₄
chromium(III) nitrate: Cr (NO₃)₃

3) Chemical equation (balanced):

4Cr + 3Pb(NO₃)₄ →  4Cr(NO₃)₃ + 3Pb

4) Mole ratios

4 mol Cr : 3 mol Pb(NO₃)₄ :  4 mol Cr(NO₃)₃ : 3 mol Pb

5) Proportion

  3 mol Pb(NO₃)₄          0.85 mol Pb(NO₃)₄
=   
      4Cr(NO₃)₃                          x

6) Solve for x:

x = 0.85 × 4 / 3 mol Pb (NO₃)₄ = 1,133 mol Pb(NO₃)₄

H tricky one let me think and get back to u