4.22g
1. Work out the equation and balance
2AgNo3 +BaCl2 ---> 2AgCl + Ba(No3)2
2. Work out formula mass of silver nitrate (should = 169.9)
3. Calculate the number of moles by doing moles = mass / formula mass so 5 divided by 169.9 = 0.0294 moles.
4. Check ratio - here the ratio is 2:2 i.e. 2 moles of silver nitrate to 2 moles of silver chloride so the moles will be the same so the moles of silver chloride is also 0.0294
5. Work out the formula mass of AgCl (always ignore big numbers at the start when working out formula mass) = 143.4
6. Work out mass by doing equation moles = mass/formula mass so
mass = moles x formula mass
mass = 0.0294 x 143.4
mass = 4.22g (3sf or 2dp)
Answer:
C.
Explanation:
if I am wrong I am so sorry
Answer:
1.79 mol.
Explanation:
- For the balanced reaction:
<em>2NaCl + F₂ → 2NaF + Cl₂.
</em>
It is clear that 2 mol of NaCl react with 1 mol of F₂ to produce 2 mol of NaF and 1 mol of Cl₂.
- Firstly, we can get the no. of moles of F₂ gas using the general law of ideal gas: <em>PV = nRT.</em>
where, P is the pressure of the gas in atm (P = 1.2 atm).
V is the volume of the gas in L (V = 18.3 L).
n is the no. of moles of the gas in mol (n = ??? mol).
R is the general gas constant (R = 0.0821 L.atm/mol.K),
T is the temperature of the gas in K (299 K).
∴ no. of moles of F₂ (n) = PV/RT = (1.2 atm)(18.3 L)/(0.0821 L.atm/mol.K)(299 K) = 0.895 mol.
- Now, we can find the no. of moles of NaCl is needed to react with 0.895 mol of F₂:
<em><u>Using cross multiplication:</u></em>
2 mol of NaCl is needed to react with → 1 mol of F₂, from stichiometry.
??? mol of NaCl is needed to react with → 0.895 mol of F₂.
∴ The no. of moles of NaCl needed = (2 mol)(0.895 mol)/(1 mol) = 1.79 mol.
The common neutralization reaction that involve NaOH reacting with HNO3 produces
NaNO3 and H2O
The equation for reaction is as folows
NaOH + HNO3 = NaNO3 + H2O
that is 1 mole of NaOH reacted with 1 mole of HNO3 to form 1 mole of NaNO3 and 1 mole of H2O
