4.) How many grams of N2 are required to react with 2.30 moles of Mg in the proc

ID: 878296 • Letter: 4

Question

4.)How many grams of N2 are required to react with 2.30 moles of Mg in the process

3 Mg + N2 --> Mg3N2 ? (Mg = 24.3 g/mol, N = 14.0 g/mol)

5.)What volume of 6.0 M sulfuric acid is required for the preparation of 500.0 mL of 0.30 M solution?

6.)What is the molarity of methanol, CH3OH (d = 0.792 g/mL) if 150.0 mL is dissolved in enough water to make 4.0 L of solution?

7.)What is the percent yield if 122 grams of SiO2 are made from 246 g of Cr2O3 by the following equation?

3 Si(s) + 2 Cr2O3(s) --> 3 SiO2(s) + 4 Cr(l)

4) 3 equivalents of Mg reacts with 1 equivalents of N2

moles of Mg = 2.30 moles

Thus, grams of N2 reacting would be = (2.30 mol / 3) x 14 g/mol = 10.73 g

5) M1V1 = M2V2

6 x V1 = 500 x 0.3

calculate V1 = 25 ml

thus, 25 ml of 6 M sulfuric acid is required for the preparation of 500.0 mL of 0.30 M solution.

6) molarity = moles/L

moles = g/molar mass

grams = density x volume = 0.792 g/ml x 150 ml = 118.8 g

molar mass of methanol = 30.04 g/mol

Feed values,

molarity = 118.8 / 30.04 x 4 = 0.99 M

7) Looking at the given equation,

2 moles of Cr2O3 forms 3 moles of SiO2

that is,

2 x 151.99 g of Cr2O3 gives 3 x 60.08 g of SiO2

thus,

246 g of Cr2O3 will give = 3 x 60.08 x 246 / 2 x 151.99 = 145.862 g of SiO2 (100% yield)

Percent yield = (actual yield / theoretical 100% yield) x 100

= (122 g / 145.862 g) x 100

= 83.64 %

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