If a solution containing
55.42
g
55.42 g
of mercury(II) nitrate is allowed to react completely with a solution containing
16.642
g
16.642 g
of sodium sulfate according to the equation below.
Hg
(
NO
3
)
2
(
aq
)
+
Na
2
SO
4
(
aq
)
⟶
2
NaNO
3
(
aq
)
+
H

Hg(NO₃)₂(aq) + Na₂SO₄(aq)   →   2NaNO₃(aq) + HgSO₄(s)

Moles of Hg(NO₃)₂ = 55.42 / 324.7 ==> 0.1707 moles

Moles of Na₂SO₄ = 16.642 / 142.04 ==> 0.1172 moles

Limiting reagent is Na₂SO₄ as it controls product formation

Moles of HgSO₄ formed = 0.1172 moles

                                        = 0.1172 x 296.65

                                        = 34.757g

Explanation:

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'

i do not get it   at all