Chemistry, 03.08.2019 14:30, devontemiles1174

What mass of carbon dioxide gas would be produced if 10g of calcium carbonate reacted with an excess of hydrochloric acid?

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Answer is: 4,4 grams of carbon dioxide gas would be produced.
Chemical reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O.
m(CaCO₃) = 10 g.
n(CaCO₃) = 10 g ÷ 100 g/mol.
n(CaCO₃) = 0,1 mol.
From chemical reaction: n(CaCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0,1 mol.
m(CO₂) = n(CO₂) · M(CO₂).
m(CO₂) = 0,1 mol· 44 g/mol.
m(CO₂) = 4,4 g.

answered: Guest

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.

answered: Guest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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