Determine the theoretical yield of HCI if 60.0 g of BCI_s and 37.5 g of H_2O are

Question

Determine the theoretical yield of HCI if 60.0 g of BCI_s and 37.5 g of H_2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCL_2 = 117.16 g mol BCI_(g) + 3H_2O_(aq) rightarrow H_2BO_2 + 3 HCI_(g) A. 75.9 g HCI B. 132 g HCI C. 187 g HCI D. 56.0 g HCI E. 25.3 g HCI The limiting reagent in a chemical reaction is the reactant A. with the smallest molar mass B. that is consumed first C. that is present in excess D. with the smallest coefficient E. none of these At constant pressure, the temperature of a closed 1.00 L sample of an ideal gas is increased from 10 degree F to 200 degree F. What is the new volume of the gas sample? A. 2.00 L B. 1.26 L C. 1.18 L D. 1.00 L E. 0.50 L A 6.35 L sample of carbon monoxide is collected at 328 K and 0.892 atm. Assuring ideal gas behavior, what volume will the gas occupy at 1.05 atm and 293 K? a. 1.96 L B. 5.48 L C. 4.82 L D. 6.10 L E. none of these Which of the following is a mathematical statement of Charles Law? A. V_1 - V_2 = T_1 - T_2 V_2/V_2 = T_1/T_2 B. V_1/T_2 = V_2/T_1, C. V_1 - T_1 = V_2/T_2 D. none of these

Explanation / Answer

Question 12

BCl3 + 3 H2O ===> H3BO3 + 3 HCl

60 gm of BCl3 = 60 /117.17 = 0.512 Mole

37.5 gm of H2O = 37.5 /18 = 2.081 moles

Limiting reagent is BCl3

0.512 Moles of BCl3 will produce 1.536 Moles of HCl

1.536 Moles of HCl = 1.536 x 36.45 = 56.01 gm

Option D is the answer

Question 13

Option B is the answer That is the consumed first

Question 14

Option C is the answer

100 F = 310.928 K

200 F = 366.48 K

Use this equation V1 /T1 =V2 /T2

v2 = 1.18

Question 15

PV=nRT

n = 6.35 x 0.892 /0.0821 x 328 = 0.21 Moles

Find the volume using the same equation

V = 0.21 x 0.0821 x 293 /1.05

V = 4.81 Liter

OPtion C is the answer.

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