Magnesium reacts with sulfuric acid to produce magnesium sulfate and hydrogen. I

Question

Magnesium reacts with sulfuric acid to produce magnesium sulfate and hydrogen. If we start with 1.4 grams of magnesium and 8.1 grams of sulfuric acid: a. How many of hydrogen can be produced? b. If the actual yield of hydrogen is 0.600 grams what is the percent yield? C. How many grams of excess reagent are remaining at the end of the reaction? When 75.0 grams of grams of ammonia, 90.0 grams of carbon dioxide and 61.0 grams of water are reacted to form ammonium carbonate, what is the limiting reactants and how many grams of product will be formed?

Explanation / Answer

Solution

Mg(s) + H2SO4 (aq) ----> MgSO4(aq) + H2(g)

Moles of Magnesium = 1.4/ 24.3 = 0.0576 moles

moles of H2SO4 = 8.1/98= 0.0826 moles

Now

a) Moles of H2 produced = 0.0576

Mass = 0.0576*2 = 0.1152 grams

b) Percentage Yield = 0.1152/0.06 = 1.92 = 192%

c) Excess Reagent ( H2SO4) = 0.0826-0.0576 = 0.025

Mass = 0.025*98= 2.45 grams

d) NH3 = 25grams = 1.47 moles

CO2= 90g = 2.045 moles

H2O= 61g = 3.388 moles

2NH3 + CO2 + H2O = (NH4)2CO3

Limiting Reagent is NH3

Moles of Product formed = 1.47/2 = 0.735 moles

Mass = 0.735*96 = 70.56 grams

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