Login

Welcome, Guest. Please login or register.

March 29, 2024, 10:29:17 am

Author Topic: stoichiometry  (Read 602 times)  Share 

0 Members and 1 Guest are viewing this topic.

hanyi

  • Fresh Poster
  • *
  • Posts: 1
  • Respect: 0
stoichiometry
« on: March 24, 2019, 07:39:10 pm »
0
calculate the mass of sodium carbonate (Na2CO3.10H2O) required to make 250mL of a 0.100 M solution.

thank you !!!
« Last Edit: March 24, 2019, 07:41:27 pm by hanyi »

fun_jirachi

  • MOTM: AUG 18
  • HSC Moderator
  • Part of the furniture
  • *****
  • Posts: 1068
  • All doom and Gloom.
  • Respect: +710
Re: stoichiometry
« Reply #1 on: March 24, 2019, 09:24:22 pm »
+2
Using c=n/V, we have that 250mL of a 0.1M solution has 0.0250 moles of Na2CO3. Finding that the molar mass of Na2CO3 is 105.988 g/mol, multiply the two together to get about 2.65g (3SF).

Hope this helps :)
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Guide Links:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)
Asking good questions