Question

A) Balance the equation: Cr(s) + SnCl4(l) ---> CrCl3(s) + Sn(s)

B) If 6.000 g of chromium and 23.000 g of tin(IV) chloride react according to this equation, whichof these is the limiting reactant?

C) What mass (in g) of chromium (III) chloride will be made?

D) What mass (in g) of the non-limiting reactant will remain after the reaction is over?

Answer 1

A)

B) Since

4 moles of Cr reacts with 3 moles of SnCl4

this means

208gm of Cr reacts with 781 g of SnCl4

therefore

6gm of Cr will react with (781/208)*6 g of SnCl4

6gm of Cr will react with 22.53g of SnCl4

Since Cr is consumed first while there is still some SnCl4 is left, Chromium is the limitng reagent.

C)

4 moles of Cr gives 4 moles of CrCl3

this means

208gm of Cr will give 633.4 g of CrCl3

therefore

6g of Cr will give (633.4/208)* 6 g of CrCl3

=> 6g of Cr will give 18.27 g of CrCl3

D) After the reaction is over

SnCl4 left=23-22.53=0.47g

=>SnCl4 left0.47g