How many milliliters of 3.0 M NaOH are required to react with 4.0 mL of 16.0 M H

Question

How many milliliters of 3.0 M NaOH are required to react with 4.0 mL of 16.0 M HNO_3? How many milliliters of 3.0 M NaOH are required to react with 0.50 g of Cu^2+ to form Cu(OH)_2? What is the theoretical percentage copper in Cu(OH)_2? What theoretical weight of CuO can be obtained from 2.00 g of Cu? Iron is obtained from iron ore according to the following reaction. Fe_2 O_3 + 3 CO rightarrow 2 Fe + 3 CO_2 (unbalanced) Assuming the blast furnace is 90.0% efficient in recovering the iron, what is the actual mass of iron obtainable from a ton of ore? 1 ton = 2000.0 lb 1 lb = 453.6 g

Explanation / Answer

The reaction is HNO3+NaOH-------> NaNO3+H2O

1 mole of HNO3 requires 1 mole of NaOH

hence moles of HNO3= 16*4 mlM

moles of NaOH also wlll be 16*4 mlM

volume of NaOH= 16*4/3= 21.33 ml

2. Cu+2+2NaOH ------->Cu(OH)2 + 2Na+

1 mole of Cu+2 requires 2 moles of NaOH

moles of Cu+2= 0.5/63.5 =0.0078

moles of NaOH= 2*0.0078= 0.0156 moles

L of NaOH = 0.0156/3 =0.0052 L =5.2 ml

3. Cu(OH)2 molar mass = 63.5+34= 97.5

mass % of Cu= 100*63.5/97.5=65.13

3 Cu+1/2O2----->CuO

molar masses : Cu= 63.5, O2= 16, CuO= 79.5

moles of Cu= 2/63.5 =0.0315

mass =moles* molar mass

moles CuO = moles of Cu= 0.0315 mass of CuO obtained = 0.0315*79.5 =2.50425

4. The balacned reaction is Fe2O3 +3CO -------> 2Fe +3CO2

molar masses : Fe2O3= 160, CO=30, Fe= 56, CO2=44

1 tons of Fe2O3 correspond to =1000 kg of ore

moles of Fe2O3 = 1000/160 =6.25 kgmoles of Fe formed = 2* moles of Fe2O3= 2*6.25= 12.5

when the furnace is 90% efficient, moles of Fe formed = 12.5*0.9=11.25

Mass of Fe formed= 11.25*56=630 kg

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