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OLga [1]
3 years ago
7

A sample of sodium reacts completely with 0.213 kg of chlorine, forming 351 g of sodium chloride. What mass of sodium reacted?

Chemistry
1 answer:
Over [174]3 years ago
8 0
Sodium reacts with chlorine as shown in the equation below; 
2Na(s) + Cl2(g) = 2NaCl(s)
The molar mass of sodium chloride is 58.44g/mol,
while that of sodium is 23 g/mol.
The moles of sodium cholride formed will be 351 g / 58.44 g = 6.006 moles
The mole ratio of sodium to Sodium chloride is 2 : 2,
therefore the moles of sodium used will be 6.006 moles,
Hence, mass of sodium used, 6.006 × 23 = 138.142 g or 0.1381 kg
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katrin [286]

Answer:

\boxed{\text{2, 1, 4, 2, 1, 2}}

Explanation:

NO₃⁻ + Sn²⁺ + __ → NO₂ + Sn⁴⁺ + __

Step 1: Separate into two half-reactions.

NO₃⁻ ⟶ NO₂

Sn²⁺ ⟶ Sn⁴⁺

Step 2: Balance all atoms other than H and O.

Done

Step 3: Balance O.

NO₃⁻ ⟶ NO₂ + H₂O

Sn²⁺ ⟶ Sn⁴⁺

Step 4: Balance H

NO₃⁻ + 2H⁺ ⟶ NO₂ + H₂O

Sn²⁺ ⟶ Sn⁴⁺

Step 5: Balance charge.

NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O

Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻

Step 6: Equalize electrons transferred.

2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]

1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]

Step 7: Add the two half-reactions.

2 × [NO₃⁻ + 2H⁺ + e⁻ ⟶ NO₂ + H₂O]

<u>1 × [Sn²⁺ ⟶ Sn⁴⁺ + 2e⁻]                                          </u>

     2NO₃⁻ + Sn²⁺ + 4H⁺ ⟶ 2NO₂ + Sn⁴⁺ + 2H₂O

Step 8: Check mass balance.

 On the left: 2 N, 6 O, 1 Sn, 4H

On the right: 2 N, 6 O, 1 Sn, 4H

Step 9: Check charge balance.

 On the left: -2 + 6 = +4

On the right: +4

The equation is balanced.

\text{The coefficients are }\boxed{\textbf{2, 1, 4, 2, 1, 2}}

4 0
4 years ago
a man bought two watches for Rs 2500 each.He sold one of them at 10% . profit and the other at 4% loss.what was his gain or loss
timofeeve [1]

Answer:

Selling price of watch A and watch B= Rs. 425

Let cost price of watch A=x

Profit =10%

∴

100

110

x=425

x=386.36

Let cost price of watch B=y

Loss =10%

∴

100

90

y=425

y=472.22

Net cost price =x+y=858.5858

Net selling price =850

Loss%=

858.58

858.5858−850

×100=1%

Explanation:

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