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frosja888 [35]
3 years ago
5

If 57.0 g of b2o3 is added to 44.7 g of cl2 and 68.8 g of c, what is the theoretical yield of boron trichloride?

Chemistry
1 answer:
sdas [7]3 years ago
6 0
(57.0 g B2O3 / (69.6202 g B2O3/mol) x (4mol BCI3 / 2 mol B2O3) = 1.64 mol BC13

(44.7 g C12) / (70.9064 g C12/mol) x (4mol BCI3 / 6mol C12) = 0.42027 mol BC13

(68.8 g C) / (12.01078 G C/mol) x (4mol BCI3 / 3 mol C) = 7.63 mol BCI3

C12 is the limiting reactant.

(0.42027 mol BCI3) X (117 . 170 g BCI3/mol) = 49.2 g BCI3 in theory.
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3 years ago
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How many grams of NaCl (molecular weight = 58 Da) would you have to dissolve in 50ml to make a 2M stock solution? Answer:
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Answer:

5.8 g

Explanation:

Molecular weight in Daltons is equivalent to the molecular weight in grams per mole.

The amount of NaCl required is calculated as follows:

(2 mol/L)(50 mL)(1 L/1000 mL) = 0.1 mol

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Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
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90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

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4 years ago
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What are equation for the chemical change that produces water from two hydrogen molecules and one oxygen molecule. And the react
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Answer:

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products: H2O

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