Asked by Jessica
If 0.605 g of magnesium hydroxide reacts with 1.00g of sulfuric acid, what is the mass of magnesium sulfate produced?
Mg(OH)2(s)+H2SO4(l)→MgSO4(s)+H2O(l)
.605g Mg x 1molMgSO/1molMg=.605g MgSo
1.00g MgSO/1mol H2SO4=1.00g MgSO
.605g+1.00g=1.605g
This is incorrect, but I don't understand what else to do.
Mg(OH)2(s)+H2SO4(l)→MgSO4(s)+H2O(l)
.605g Mg x 1molMgSO/1molMg=.605g MgSo
1.00g MgSO/1mol H2SO4=1.00g MgSO
.605g+1.00g=1.605g
This is incorrect, but I don't understand what else to do.
Answers
Answered by
Damon
molar mass of Mg(OH)2 = 24.3+2*17 = 58.3 g
so we have
.605/58.3 = .0104 mol
molar mass of H2SO4 = 2+32+64 = 98 g
so we have
1/98 = .0102 mol
so
we will get .0102 mol of MgSO4
.0102 ( 24.3+32+64) = 1.23 g
so we have
.605/58.3 = .0104 mol
molar mass of H2SO4 = 2+32+64 = 98 g
so we have
1/98 = .0102 mol
so
we will get .0102 mol of MgSO4
.0102 ( 24.3+32+64) = 1.23 g
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