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Assuming an efficiency of 38.30%, 38.30%, calculate the actual yield of magnesium nitrate formed from 114.4 g 114.4 g of magnesium and excess copper(II) nitrate. Mg+Cu(NO3)2⟶Mg(NO)2+Cu
Mg + Cu(NO3)2 = Mg(NO3)2 + CuMg = Mg(NO3)2 24 g Mg = 148g Mg(NO3)2 1g Mg = 148/24 Mg(NO3)2 114.4g Mg = 148× 114.4g/ 24 Mg(NO3)2 = 705.5g (...
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If 10.00 g of magnesium is reacted with 95.75 g of copper (II) sulphate, magnesium sulphate and copper are formed.
a. Which reactant is in excess?
b. Calculate the mass of copper formed.
Mg + CuSO4 = MgSO4 + Cu n(Mg) = m/M M(Mg) = 24 g/mol n(Mg) = 10g/24g/mol = 0.417 mol n(CuSO4) = m/M M(CuSO4)= 64 + 32 + 16*4 =...
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If 10.45 g of aluminum are reacted with 66.55 g of copper (II) sulphate, aluminum sulphate and copper are formed.
a.Which reactant is in excess?
b. Calculate the mass of copper formed.
Balanced reaction equation:2Al + 3CuSO4 = 3Cu + Al2(SO4)3Molar weight of Al: 26.98 g/molQuantity of Al: 10.45 g / 26.98 g/mol = 0.387 molMolar weight of CuSO4: 159.61 g/molQuantity of...
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What is the theoretical yield (in grams) of C9H8O4 when 2.00 g C7H6O3 is heated
with 4.00 g of C4H6O3? If the actual yield is 1.98 g, what is the percent yield?
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2
C7H6O3 + C4H6O3 rarr; C9H8O4 + C2H4O2 n of C7H6O3 = 2/ (12*7 + 6 + 3*16) = 0.0145 moles this compound is less n of C4H6O3 = 4/(12*4 +...
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Calculate the percentage yield of this reaction of 2MgO(s) if 0,15 moles of magnesium react with excess oxygen gas to form 5,4 grams of Magnesium oxide
Given Moles of Mgo is 0.15 moles . Mg react with O2 and form MgO .2 mole of Mg gives 2 mole of Mgo .So 0.15 moles of Mg Gives...
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Balance equation of : 20cm^3 of 0.5 mol dm^-3 hydrocloric acid + excess of magnesium powder and 20cm^3 of 0.5 mol dm^-3 sulphric acid + excess of magnesium powder.
2HCl+Mg=MgCl2+H2H2SO4+Mg=MgSO4+H2
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combining 0.385 mol of feo3 with excess carbon produced 16.5g of fe.
what is the actual yield of iron in moles?
what was the theoretical yield of iron in moles?
what was the percent yield?
.n (FeO3)=0.385 mol .m (Fe)=16.5 g .n(Fe)theor=? %=? FeO3+ 3C=Fe+3CO .n (Fe)theor= n (FeO3)=0.385 mol Amount of substance of iron n (Fe)=m/Ar= 16.5/56=0.295 mol %= n(Fe)/ n (Fe) theor= 0.295/0.385...
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When 250 kg of copper (I) oxide is heated with 129 kg of copper (I) sulfide, 285 kg of copper is recovered. Calculate the theoretical yield of copper.
2Cu2O + Cu2S -> 6Cu + SO2n= m/Mn(Cu2O)= 250000g/143.09g/mol= 1747 moln(Cu2S)= 129000g/ 159.16 g/mol= 810.5 molFind limiting reactant: compare1747/2 and 810.5873.5>810.5, consequently , Cu2S is a limiting reactant.According to equation...
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What is my actual yield of H2SO4 if I had a theoretical yield of 76 grams of H2SO4 and a percent yield of 79%?
76 x 0.79 = 60.04 g
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