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DedPeter [7]
3 years ago
5

How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 15.0 g of hydrochloric aci

d?
Chemistry
1 answer:
horsena [70]3 years ago
5 0

Answer:

22.7 g of CaCl₂ are produced in the reaction

Explanation:

This is the reaction:

CaCO₃ +  2HCl  →  CaCl₂  + CO₂  +  H₂O

Now, let's determine the limiting reactant.

Let's divide the mass between the molar mass, to find out moles of each reactant.

29 g / 100.08 g/m = 0.289 of carbonate

15 g / 36.45 g/m = 0.411 of acid

1 mol of carbonate must react with 2 moles of acid

0.289 moles of carbonate will react with the double of moles (0.578)

I only have 0.411 of HCl, so the acid is the limiting reactant.

Ratio is 2:1, so I will produce the half of moles, of salt.

0.411 / 2 = 0.205 moles of CaCl₂

Mol . molar mass = mass → 0.205 m . 110.98 g/m = 22.7 g

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Why do acids and bases neutralize eachother?
timofeeve [1]

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3 years ago
What is the density of an 820 g sample of pure silicon occupying a 350 cm3 container
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Answer:

2.343 g/ cm³

Explanation:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Given data:

density = ?

volume= 350 cm³

mass= 820 g

Now we will put the values in the formula,

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6 0
3 years ago
A 12400. mL container holds a sample of argon gas at 35.00C and
xz_007 [3.2K]

Answer:

0.574moles

Explanation:

Using the general gas equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (Litres)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (Kelvin)

According to the information provided in the question;

- Volume (V) = 12400mL = 12400/1000 = 12.4L

- Pressure (P) = 890mmHg = 890/760 = 1.17atm

- Temperature (T) = 35°C = 35 + 273 = 308K

Hence, using PV = nRT

n = PV/RT

n = 1.17 × 12.4 ÷ 0.0821 × 308

n = 14.508 ÷ 25.287

n = 0.574moles

Therefore, the number of moles of argon gas in the cylinder is 0.574moles

7 0
3 years ago
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