Question

Какой объем углекислого газа (н.у.) выделится при взаимодействии 15 г карбоната натрия, содержащего 15% примесей, с соляной кислотой?

Answer 1

Na2CO3+2HCl=2NaCl+CO2+H2O

m(Na2CO3)=15-15*0/15=12.75 г

n(Na2CO3)=n(CO2)=m/M=12.75/106=0.12 моль

V(CO2)=0.12*22.4=2.688 л 

Answer 2

I can chat with you in Russian and help you with various tasks. You asked me to answer in detail how much carbon dioxide gas (STP) will be released when 15 g of sodium carbonate containing 15% impurities reacts with hydrochloric acid. Here is my answer:

The reaction equation is:

$$Na_2CO_3 + 2HCl \rightarrow 2NaCl + CO_2 + H_2O$$

The molar mass of sodium carbonate is 106 g/mol. The mass of pure sodium carbonate in 15 g of the mixture is 85% of 15 g, which is 12.75 g. The number of moles of sodium carbonate is:

$$n_{Na_2CO_3} = \frac{m_{Na_2CO_3}}{M_{Na_2CO_3}} = \frac{12.75}{106} = 0.12 \text{ mol}$$

According to the reaction equation, one mole of sodium carbonate produces one mole of carbon dioxide. Therefore, the number of moles of carbon dioxide is equal to the number of moles of sodium carbonate, which is 0.12 mol. The volume of carbon dioxide at STP is:

$$V_{CO_2} = n_{CO_2} \times V_m = 0.12 \times 22.4 = 2.69 \text{ L}$$

where $V_m$ is the molar volume of an ideal gas at STP, which is 22.4 L/mol.

Therefore, the volume of carbon dioxide gas (STP) that will be released when 15 g of sodium carbonate containing 15% impurities reacts with hydrochloric acid is 2.69 L.Ответь+подробно.+Какой+объем+углекислого+газа+(н.у.)+выделится+при+взаимодействии+15+г+карбоната+натрия,+содержащего+15%+примесей,+с+соляной+кислотой?