The electronegativity value for Mg is 1.2 and the value for O is 3.5. Based on t

Question

The electronegativity value for Mg is 1.2 and the value for O is 3.5. Based on these values, what type of bond is expected for a compound formed between Mg and O? Polar covalent Pure covalent Ionic not enough information The theoretical yield of a reaction 750 grams of product and the actual yield is 56.25g. What is the percent yield? 31.5 75.0 56.0 178 Consider the following equation: CO+2H_2 CH_3OH delta H r times m = -128 kJ Calculate the amount of heat (in kJ) associated with complete reaction of 8.08 g H_2 -512 kJ -256 kJ -1034 kJ -1024 kJ Given that 4 NH_3 + 5O_2 - 4NO + 6H_2O, when 4.50 mol of H_2O are formed, the NO formed is 4.500 mol 3.00 mol. 1.50 mol. 6.75 mol. How many moles of H_2 can be made from complete reaction 60moles of Al? Given: 2 Al + 6 HCl-2 AlCl_3 + 3H_2 9.0 moles 18 moles 3.0 moles 4.5 moles

Explanation / Answer

1. The electronegativity difference between Mg (1.2) & O (3.5) is (3.5-1.2 = 2.3) 2.3.

The electronegativity difference is much higher making the electrons to shift towards one atom. As this difference is much higher the electrons will be shifted towards one atom; making it a IONIC bond.

2. Percent yield = ( Experimental yield / Theoretical yield ) * 100 = ( 56.25 / 75 ) *100 = 75%

3. CO + 2H2 --> CH3OH .... dH = -128kJ

For this reaction, one mole of CO reacts with 2 moles of H2 to form one mole of CH3OH.

8.08 gm of H2 = 8.08 gms / 2.004 gms/mole = 4 moles.

So 4 moles of H2 will reaction with 2 moles of CO to form 4 moles of CH3OH; librating 2x times dH. This will be equal to (128*2 = 256) 256kJ.

4. 4NH3 + 5O2 --> 4NO + 6H2O.

In this reaction 6 moles of H2O are formed for every 4 moles of NO formed.

If 4.5 moles of H2O are formed; then it will be ((4.5 * 4)/ 6 = 3 ) 3 moles of NO.

5. 2Al + 6HCl ---> 2 AlCl3 + 3H2

In this reaction 3 moles of H2 are formed for every 2 moles of Al consumed completely.

So if we consume 6 moles of Al; then we will form 9 moles of H2.

Please note that for any reaction balance; MOLES of the reactants & products are considered as the basis for calculation.

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