Question

Какая масса соли получится при реакции с оксидом цинка 2 литров этановой кислоты, если выход продукта равен 80 %

Answer 1

2CH3COOH+ZnO=(CH3COO)2Zn + H2O

u=2 моль                      u=1 моль

Mr=22,4                       Mr=(12*2+3+16*2)*2+65=183 грамма

m==22,4*2=44,8     

2/44,8=x/183

x=183*2/44.8

x=8,16 грамм

w(выхода)=w(практического выхода)/w(теоретического выхода)

0,80=x/8,16

x=0,80*8,16

x=6,528 грамм

Answer 2

Reaction and Product Yield

The reaction of zinc oxide with ethanoic (acetic) acid can be represented by the following chemical equation: ZnO + CH3COOH → Zn(CH3COO)2 + H2O

Given that the yield of the product is 80%, we can calculate the mass of salt produced using the following steps:

1. Calculate the molar mass of the reactants and products. 2. Use the molar ratio from the balanced chemical equation to find the moles of the product. 3. Convert moles of the product to mass using the molar mass of the product.

Calculation Steps

1. Molar Mass Calculation: - Molar mass of ZnO (Zinc Oxide) = 81.38 g/mol - Molar mass of CH3COOH (Ethanoic Acid) = 60.05 g/mol - Molar mass of Zn(CH3COO)2 (Zinc Acetate) = 183.43 g/mol

2. Moles of the Product: - From the balanced chemical equation, 1 mole of ZnO reacts with 2 moles of CH3COOH to produce 1 mole of Zn(CH3COO)2. - Using the molar ratio, calculate the moles of Zn(CH3COO)2 produced.

3. Mass of the Salt: - Convert the moles of Zn(CH3COO)2 to mass using its molar mass.

Calculation

Using the given information, the mass of the salt produced can be calculated as follows:

1. Moles of Zn(CH3COO)2 produced: - Moles of Zn(CH3COO)2 = (2 L * 0.80) / 60.05 g/mol - Moles of Zn(CH3COO)2 = 0.0266 moles

2. Mass of Zn(CH3COO)2 produced: - Mass of Zn(CH3COO)2 = 0.0266 moles * 183.43 g/mol - Mass of Zn(CH3COO)2 = 4.88 grams

Therefore, the mass of the salt produced in the reaction is approximately 4.88 grams.