#1 is worded so poorly I don't understand it. My biggest problem is that I don't know what volume the CO2 is dissolved in; e.g., is CO2 dissolved in 920 or in 1800 or is it 920+1800 mL?
#2.
Determine mols MnS by 1.15/87.00 = ??
Determine mols HCl by M x L = 0.800 x 0.0288 = ??
Determine which is the limiting reagent. I expect it is HCl (at least I'm assuming that).
Determine how much of the HCl will react with the MnS. It will be 1/2 of the mols HCl, then subtract that from mols MnS initially and multiply the difference by 87 to obtain the mass MnS remaining. I ignored any H^+ produced by H2S. Check my thinking. Check my typing for typos.
Ive been trying to do my chemistry homework for the past few hours and I keep getting some of the questions wrong. These are the 2 questions i am having trouble with.
1.
If 0.582 L of gaseous CO2 measured at STP dissolved in 920 mL of water is produced stoichiometrically according to the balanced equation in a reaction solution with a total volume of 1800 mL, what is the molarity (M) of Na+ produced?
Na2CO3(aq) + 2 HCl(aq) �¨ CO2(g) + 2 NaCl(aq) + H2O(l)
Molar Mass (g)
CO2 44.010
Na+ 22.989
Molar Volume (L)
22.400 at STP
Gas Constant
0.0821
2.If 1.15 g of solid MnS and 800 mL of 0.0288 M aqueous HCl are reacted stoichiometrically according to the equation, how many mol of solid MnS remained?
MnS(s) + 2HCl(aq) �¨ H2S(g) + MnCl2(aq)
Molar Mass (g)
HCl 36.461
MnS 87.00
Molar Volume (L)
22.400 at STP
Gas Constant
0.0821
1 answer