Question

Suppose you use 0.0150 liter of a 2.50 M solution of silver nitrate. Assuming the reaction goes completion how much silver sulfide is produced

Answer 1

M= Moles/ L

2.50=x/0.0150

x=0.0375 moles

moles x mm= grams

0.0375 x 247.8= 9.2925 grams

Answer 2

the answer is                                                                                                            

C. 9.30 g