How many moles of zncl2 can be prepared from the reaction of 1.14 grams of zinc with excess HCI? Zn+2HCI=ZnCl2+H2

Question

asked Nov 28, 2023 122k views

1 Answer

ZayedUpal · Answer 1 · 2023-12-01T16:34:48+0000

Answer: 0.02 mol

Step-by-step explanation:

The atomic mass of zinc is 65.409 g/mol, so 1.14 grams of zinc is about
(1.14)/(65.409)=0.0174287942 moles.

From the equation, we know that for every mole of zinc consumed, there is 1 mole of zinc(II) chloride produced, so about 0.02 moles can be prepared.