Verified answer

Ответ:

дано

m(ppa HCL) = 160 g

W(HCL) = 4%

+ Na2CO3

--------------------

V(CO2) - ?

m(HCL) = 160 * 4% / 100% = 6.4 g

2HCL + Na2CO3-->2NaCL + H2O + CO2

M(HCL) = 36.5 g/mol

n(HCL) = m/M = 6.4 / 36.5 = 0.18 mol

2n(HCL) = n(CO2)

n(CO2) = 0.18 / 2 = 0.09 mol

V(CO2) = n(CO2) * Vm = 0.09 * 22.4 = 2.016 L

ответ 2.016 л

Объяснение: